Chapter 10: Conic Sections

Let the centre of the required circle be denoted by $$O(0,0)$$ and let its radius be the positive real number $$r$$.

Choose an arbitrary point $$P(x,y)$$ on the circle. By definition of a circle, the distance between the point and the centre equals the radius:

$$OP = r$$

Using the distance formula between two points $$O(0,0)$$ and $$P(x,y)$$ in the Cartesian plane, we have

$$OP = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$$

Equating this expression to the radius and then squaring both sides removes the square root:

\[ \sqrt{x^2 + y^2} = r \quad \Longrightarrow \quad x^2 + y^2 = r^2 \]

The equation

$$x^2 + y^2 = r^2$$

contains exactly the set of all points $$P(x,y)$$ whose distance from $$O(0,0)$$ is $$r$$, and hence represents the required circle.

For a circle, the standard form of the equation is

$$ (x - h)^{2} + (y - k)^{2} = r^{2} $$

where $$ (h, k) $$ is the centre and $$ r $$ is the radius.

1. Identify the given data:

   • Centre $$ (h, k) = (-3, 2) $$ ⇒ $$ h = -3,\ k = 2 $$
   • Radius $$ r = 4 $$

2. Substitute these values into the standard form:

   $$ (x - (-3))^{2} + (y - 2)^{2} = 4^{2} $$

   Simplify each part:

   $$ (x + 3)^{2} + (y - 2)^{2} = 16 $$

3. (Optional) Expand to obtain the general form:

   $$ (x + 3)^{2} + (y - 2)^{2} = 16 $$

   $$ x^{2} + 6x + 9 + y^{2} - 4y + 4 = 16 $$

   $$ x^{2} + y^{2} + 6x - 4y - 3 = 0 $$

Thus, either form is acceptable; the simplest is the standard form shown above.

The general equation of a circle in two variables is
$$x^2 + y^2 + 2gx + 2fy + c = 0,$$
whose centre is $$(-g,",-f)$$ and radius is $$r = \sqrt{g^2 + f^2 - c}.$$

Given equation:
$$x^2 + y^2 + 8x + 10y - 8 = 0.$$

Step 1: Group x-terms and y-terms.
$$x^2 + 8x \; + \; y^2 + 10y = 8.$$

Step 2: Complete the square in each variable.

• For $$x^2 + 8x$$, add and subtract $$\left(\frac{8}{2}\right)^2 = 16.$$
• For $$y^2 + 10y$$, add and subtract $$\left(\frac{10}{2}\right)^2 = 25.$$

Insert these inside the equation (adding the same amounts to both sides keeps equality):
$$x^2 + 8x + 16 + y^2 + 10y + 25 = 8 + 16 + 25.$$

Step 3: Write as perfect squares.
Left side:
$$x^2 + 8x + 16 = (x+4)^2, \quad y^2 + 10y + 25 = (y+5)^2.$$
Right side:
$$8 + 16 + 25 = 49.$$

Thus the equation becomes
$$(x+4)^2 + (y+5)^2 = 49.$$

Step 4: Read the centre and radius.
The standard form $$ (x - h)^2 + (y - k)^2 = r^2 $$ shows centre $$ (h, k) $$ and radius $$ r $$. Here
$$h = -4, \quad k = -5, \quad r^2 = 49 \;\Rightarrow\; r = 7.$$

Therefore, the circle has
centre $$(-4,-5)$$ and radius $$7$$.

Let the required circle have centre $$C\,(h,\,k)$$ and radius $$r$$.

Since the centre lies on the line $$x+y=2$$, we at once have

$$h+k=2 \qquad (1)$$

The circle passes through the two given points

• $$A\,(2,-2)$$,
• $$B\,(3,4)$$.

Therefore

$$CA=r \quad\text{and}\quad CB=r.$$

Equating the two squared distances:

$$\begin{aligned} &(2-h)^2+(-2-k)^2 &\;=&\; (3-h)^2+(4-k)^2\\[2pt] \Rightarrow\;& (2-h)^2+(-2-k)^2-(3-h)^2-(4-k)^2 &\;=&\;0. \end{aligned}$$

Expansion gives

$$\bigl(4-4h+h^2\bigr)+\bigl(4+4k+k^2\bigr)-\bigl(9-6h+h^2\bigr)-\bigl(16-8k+k^2\bigr)=0.$$

On cancelling $$h^2$$ and $$k^2$$ and collecting like terms:

$$8-4h+4k-25+6h+8k=0\;\;\Longrightarrow\;\;2h+12k=17\;\;\Longrightarrow\;\; h+6k=\tfrac{17}{2}. \qquad (2)$$

Now solve (1) and (2) simultaneously.

From (1), $$h=2-k$$. Substitute in (2):

$$2-k+6k=\tfrac{17}{2}\;\;\Longrightarrow\;\;2+5k=\tfrac{17}{2}\;\;\Longrightarrow\;\;5k=\tfrac{13}{2}\;\;\Longrightarrow\;\;k=\tfrac{13}{10}.$$

Hence $$h=2-\tfrac{13}{10}=\tfrac{7}{10}.$$

Thus the centre is $$\bigl(\tfrac{7}{10},\,\tfrac{13}{10}\bigr).$$

Radius:

$$\begin{aligned} r^2 &= (2-h)^2+(-2-k)^2\\[2pt] &=\Bigl(2-\tfrac{7}{10}\Bigr)^2+\Bigl(-2-\tfrac{13}{10}\Bigr)^2\\[2pt] &=\Bigl(\tfrac{13}{10}\Bigr)^2+\Bigl(-\tfrac{33}{10}\Bigr)^2\\[2pt] &=\tfrac{169}{100}+\tfrac{1089}{100}=\tfrac{1258}{100}=\tfrac{629}{50}. \end{aligned}$$

Equation in centre–radius form:

$$ (x-\tfrac{7}{10})^2+(y-\tfrac{13}{10})^2=\tfrac{629}{50}. $$

For an equation without fractions multiply by $$50$$:

$$50(x-\tfrac{7}{10})^2+50(y-\tfrac{13}{10})^2=629.$$

On expanding and simplifying,

$$5x^2+5y^2-7x-13y-52=0.$$

Given data

Centre : $$C(h,k)=(0,2)$$
Radius : $$r=2$$

Standard form of a circle

With centre $$C(h,k)$$ and radius $$r$$ the equation is
$$(x-h)^{2}+(y-k)^{2}=r^{2}$$.

Insert the given values

Substituting $$h=0,\,k=2,\,r=2$$,
$$(x-0)^{2}+(y-2)^{2}=2^{2}.$$

Simplify

Since $$(x-0)^{2}=x^{2}$$ and $$2^{2}=4$$, we obtain
$$x^{2}+(y-2)^{2}=4.$$

Hence, the required equation of the circle is $$x^{2}+(y-2)^{2}=4$$.

For a circle whose centre is $$C(h, k)$$ and radius is $$r$$, the standard (centre–radius) form of the equation is

$$ (x - h)^2 + (y - k)^2 = r^2 $$

Here the given data are

• Centre: $$(-2, 3) \Rightarrow h = -2,\; k = 3$$
• Radius: $$r = 4$$

Substituting these values in the standard form:

$$ (x - (-2))^2 + (y - 3)^2 = 4^2 $$

Simplify the signs and square the radius:

$$ (x + 2)^2 + (y - 3)^2 = 16 $$

(Optional expansion to general form)

Expand each square:

$$ (x^2 + 4x + 4) + (y^2 - 6y + 9) = 16 $$

Combine like terms on the left and bring $$16$$ to that side:

$$ x^2 + y^2 + 4x - 6y + 4 + 9 - 16 = 0 $$

$$ x^2 + y^2 + 4x - 6y - 3 = 0 $$

Thus the required equation of the circle can be written either in centre–radius form or in general quadratic form.

The general (centre–radius) equation of a circle is

$$ (x-h)^2 + (y-k)^2 = r^2 $$

Here the centre is $$\bigl(h,k\bigr)=\left(\dfrac12,\dfrac14\right)$$ and the radius is $$r=\dfrac1{12}$$. Substituting these values:

$$ \bigl(x-\tfrac12\bigr)^2 + \bigl(y-\tfrac14\bigr)^2 = \left(\tfrac1{12}\right)^2 $$

$$ \bigl(x-\tfrac12\bigr)^2 + \bigl(y-\tfrac14\bigr)^2 = \dfrac1{144} $$

This is already a perfectly valid equation of the required circle.

If an expanded Cartesian form is desired, open the squares and collect like terms:

$$ (x^2 - x + \tfrac14) + (y^2 - \tfrac12 y + \tfrac1{16}) = \dfrac1{144} $$

$$ x^2 + y^2 - x - \tfrac12 y + \left(\tfrac14+\tfrac1{16}-\tfrac1{144}\right) = 0 $$

$$ x^2 + y^2 - x - \tfrac12 y + \tfrac{11}{36} = 0 $$

Hence, either form below represents the same circle:

• Centre–radius form: $$\bigl(x-\tfrac12\bigr)^2 + \bigl(y-\tfrac14\bigr)^2 = \dfrac1{144}$$
• General form: $$x^2 + y^2 - x - \tfrac12 y + \dfrac{11}{36}=0$$ (or $$36x^2+36y^2-36x-18y+11=0$$ after clearing denominators).

Step 1 – Recall the standard form
The equation of a circle with centre $$C(h,k)$$ and radius $$r$$ is
$$ (x-h)^2 + (y-k)^2 = r^2 $$.

Step 2 – Insert the given centre
Here $$h = 1$$ and $$k = 1$$, so
$$ (x-1)^2 + (y-1)^2 = r^2 $$.

Step 3 – Insert the given radius
The radius is $$r = \sqrt{2}$$, hence $$r^2 = (\sqrt{2})^2 = 2$$. Therefore
$$ (x-1)^2 + (y-1)^2 = 2 $$.

Optional Step 4 – Expanded form
Expanding the squares gives
$$ x^2 + y^2 - 2x - 2y + 1 + 1 = 2 \;\Longrightarrow\; x^2 + y^2 - 2x - 2y = 0. $$

Either form represents the required circle.

Given data

• Centre of the circle : $$(-a,-b)$$
• Radius of the circle : $$r = \sqrt{a^{2}-b^{2}}$$ (for a real radius we tacitly assume $$a^{2} \ge b^{2}$$).

Standard form of a circle

A circle whose centre is $$(h,k)$$ and radius is $$r$$ is written as

$$ (x-h)^{2} + (y-k)^{2} = r^{2}. $$

Substitute the given centre and radius

Here $$(h,k)=(-a,-b)$$ and $$r^{2}=a^{2}-b^{2}$$. Hence

$$ (x-(-a))^{2} + (y-(-b))^{2} = a^{2}-b^{2}. $$

Simplify the double negatives:

$$ (x + a)^{2} + (y + b)^{2} = a^{2} - b^{2}. $$

This is already the simplest compact form of the required circle.

Optional expansion (sometimes expected)

If you expand the left-hand side, you obtain

$$ x^{2} + 2ax + a^{2} + y^{2} + 2by + b^{2} = a^{2} - b^{2}. $$

Move all terms to the left:

$$ x^{2} + y^{2} + 2ax + 2by + b^{2} + b^{2} = 0, $$

that is,

$$ x^{2} + y^{2} + 2ax + 2by + 2b^{2} = 0. $$

Either form is acceptable; the condensed form $$ (x + a)^{2} + (y + b)^{2} = a^{2} - b^{2} $$ is usually preferred.

Step 1 : Recall the standard form of a circle.
A circle with centre $$C(h,k)$$ and radius $$r$$ is written as
$$ (x-h)^2 + (y-k)^2 = r^2 $$.

Step 2 : Compare the given equation with the standard form.
The given equation is
$$ (x+5)^2 + (y-3)^2 = 36 $$.

• The first bracket can be rewritten as $$x-(-5)$$, so $$h=-5$$.
• The second bracket already matches $$y-k$$ with $$k = 3$$.
• The right‐hand side equals $$r^2$$, hence $$r^2 = 36$$.

Step 3 : Write the centre.
Therefore, the centre is
$$ C(h,k) = (-5,3) $$.

Step 4 : Find the radius.
Take the square root of both sides of $$r^2 = 36$$:
$$ r = \sqrt{36} = 6 $$.

Conclusion.
The circle has centre $$(-5,3)$$ and radius $$6$$.

We rewrite the given general-form equation of a circle by grouping the $$x$$-terms and the $$y$$-terms:

$$x^{2}-4x+y^{2}-8y-45=0$$

Move the constant term to the right-hand side:

$$x^{2}-4x+y^{2}-8y=45$$

Step 1  Complete the square in $$x$$.
Take half of the coefficient of $$x$$ (which is $$-4$$), square it, and add it to both sides:

Half of $$-4$$ is $$-2$$, and $$(-2)^{2}=4$$, so add $$4$$:

$$x^{2}-4x+4+y^{2}-8y=45+4$$

Step 2  Complete the square in $$y$$.
Half of $$-8$$ is $$-4$$, and $$(-4)^{2}=16$$, so add $$16$$ to both sides:

$$x^{2}-4x+4+y^{2}-8y+16=45+4+16$$

Simplify each side:

$$\bigl(x^{2}-4x+4\bigr)+\bigl(y^{2}-8y+16\bigr)=65$$

Each bracket is now a perfect square trinomial:

$$\bigl(x-2\bigr)^{2}+\bigl(y-4\bigr)^{2}=65$$

This is the standard form $$\bigl(x-h\bigr)^{2}+\bigl(y-k\bigr)^{2}=r^{2}$$, where

• $$h=2$$, $$k=4$$, so the centre is $$(2,4)$$.
• $$r^{2}=65$$, so the radius is $$r=\sqrt{65}$$.

The general equation of a circle in the Cartesian plane is

$$x^{2}+y^{2}+2gx+2fy+c=0$$

Its centre is $$(-g,-f)$$ and its radius is $$\sqrt{g^{2}+f^{2}-c}$$ (provided $$g^{2}+f^{2}-c>0$$).

Compare the given equation

$$x^{2}+y^{2}-8x+10y-12=0$$

with the general form. Rewrite the linear terms so that the coefficients match $$2g$$ and $$2f$$ explicitly:

$$x^{2}+y^{2}-8x+10y-12=0\;\Longrightarrow\;x^{2}+y^{2}+2(-4)x+2(5)y-12=0$$

From this identification,

• $$2g=-8\;\Rightarrow\;g=-4$$
• $$2f=10\;\Rightarrow\;f=5$$
• $$c=-12$$

Therefore

Centre $$=(-g,-f)=(4,-5).$$

For the radius, compute $$g^{2}+f^{2}-c$$:

$$g^{2}+f^{2}-c=(-4)^{2}+5^{2}-(-12)=16+25+12=53.$$

Radius $$=\sqrt{53}.$$

As a check, complete the squares directly:

$$\begin{aligned} x^{2}+y^{2}-8x+10y-12&=0\\[2pt] (x^{2}-8x)+(y^{2}+10y)&=12\\[2pt] (x^{2}-8x+16)+(y^{2}+10y+25)&=12+16+25\\[2pt] (x-4)^{2}+(y+5)^{2}&=53, \end{aligned}$$

confirming the same centre and radius.

The given equation is

$$2x^2 + 2y^2 - x = 0.$$

Step 1   Divide every term by 2 so that the coefficients of $$x^2$$ and $$y^2$$ become 1:

$$x^2 + y^2 - \frac12 x = 0.$$

Step 2   Re-arrange by keeping the $$x$$-terms together:

$$x^2 - \frac12 x + y^2 = 0.$$

Step 3   Complete the square in $$x$$.

• The coefficient of $$x$$ is $$-\dfrac12$$.
• Half of this coefficient is $$-\dfrac14$$.
• The square of $$-\dfrac14$$ is $$\dfrac1{16}$$.

Add and subtract $$\dfrac1{16}$$ in the left side (this does not change the equality):

$$x^2 - \frac12 x + \frac1{16} - \frac1{16} + y^2 = 0.$$

Step 4   Group the perfect square and move the constant term to the right:

$$(x - \frac14)^2 + y^2 = \frac1{16}.$$

This is the standard form $$ (x - h)^2 + (y - k)^2 = r^2 $$ of a circle.
Comparing, we get

• Centre $$C(h, k) = \left( \dfrac14,\; 0 \right)$$,
• Radius $$r = \sqrt{\dfrac1{16}} = \dfrac14.$$

Thus the required centre and radius are found.

Let the centre of the required circle be $$C(h, k)$$. Because the centre lies on the given straight line,

$$4h + k = 16 \qquad (1)$$

The circle has to pass through the two points $$P(4, 1)$$ and $$Q(6, 5)$$, so their distances from $$C$$ must be equal to the radius. Hence

$$CP^2 = CQ^2$$

That is,

$$ (h - 4)^2 + (k - 1)^2 = (h - 6)^2 + (k - 5)^2 \qquad (2)$$

Expand (2):

$$h^2 - 8h + 16 + k^2 - 2k + 1 = h^2 - 12h + 36 + k^2 - 10k + 25$$

Simplify (the $$h^2$$ and $$k^2$$ terms cancel):

$$4h + 8k - 44 = 0 \;\Longrightarrow\; h + 2k = 11 \qquad (3)$$

Solve the simultaneous linear equations (1) and (3).

• From (3): $$h = 11 - 2k$$.
• Substitute in (1): $$4(11 - 2k) + k = 16 \;\Longrightarrow\; 44 - 8k + k = 16 \;\Longrightarrow\; -7k = -28 \;\Longrightarrow\; k = 4$$.
• Then $$h = 11 - 2\times4 = 3$$.

So the centre is $$C(3, 4)$$.

Radius:

$$r^2 = (3 - 4)^2 + (4 - 1)^2 = (-1)^2 + 3^2 = 1 + 9 = 10 \;\Longrightarrow\; r = \sqrt{10}$$

Equation of the circle in centre–radius form:

$$(x - 3)^2 + (y - 4)^2 = 10$$

If an expanded (general) form is desired,

$$x^2 - 6x + 9 + y^2 - 8y + 16 = 10$$

$$\Longrightarrow\; x^2 + y^2 - 6x - 8y + 15 = 0$$

Let the centre of the required circle be $$C(h,k)$$.

Because the centre lies on the given straight line,

$$h-3k-11=0 \;\;\Longrightarrow\;\; h=3k+11\;\;\;\;(1)$$

The circle passes through the two given points $$A(2,3)$$ and $$B(-1,1)$$, so their distances from the centre are equal to the radius. Hence

$$CA^{2}=CB^{2}$$

that is,

$$(h-2)^{2}+(k-3)^{2}=(h+1)^{2}+(k-1)^{2}\;\;\;\;(2)$$

Simplifying equation (2)

Expand and collect like terms:

$$(h^{2}-4h+4)+(k^{2}-6k+9)=(h^{2}+2h+1)+(k^{2}-2k+1)$$

Cancel $$h^{2}$$ and $$k^{2}$$, then bring all terms to one side:

$$(-4h-2h)+(-6k+2k)+(4+9-1-1)=0$$

$$-6h-4k+11=0 \;\;\Longrightarrow\;\; 6h+4k-11=0\;\;\;\;(3)$$

Solving the simultaneous linear equations (1) and (3)

From (1): $$h=3k+11$$. Substitute in (3):

$$6(3k+11)+4k-11=0$$

$$18k+66+4k-11=0$$

$$22k+55=0 \;\;\Longrightarrow\;\; k=-\dfrac{55}{22}=-\dfrac52$$

Then from (1):

$$h=3\left(-\dfrac52\right)+11=-\dfrac{15}{2}+\dfrac{22}{2}=\dfrac72$$

Hence the centre is $$C\bigl(\dfrac72,-\dfrac52\bigr)$$.

Finding the radius

Use point $$A(2,3)$$:

$$r^{2}=(h-2)^{2}+(k-3)^{2}=\left(\dfrac72-2\right)^{2}+\left(-\dfrac52-3\right)^{2}$$

$$=\left(\dfrac32\right)^{2}+\left(-\dfrac{11}{2}\right)^{2}=\dfrac{9}{4}+\dfrac{121}{4}=\dfrac{130}{4}=\dfrac{65}{2}$$

Equation of the circle

$$(x-h)^{2}+(y-k)^{2}=r^{2}$$

$$\bigl(x-\dfrac72\bigr)^{2}+\bigl(y+\dfrac52\bigr)^{2}=\dfrac{65}{2}$$

To remove fractions, multiply by 2 and expand:

$$2\left[\left(x-\dfrac72\right)^{2}+\left(y+\dfrac52\right)^{2}\right]=65$$

$$x^{2}+y^{2}-7x+5y-14=0$$

Therefore, the required circle is

$$x^{2}+y^{2}-7x+5y-14=0$$

Step 1 : Let the centre be on the x-axis

If the centre lies on the x-axis its coordinates can be written as $$(a,0)$$ for some real number $$a$$.

Step 2 : Use the given radius

The radius of the circle is $$5$$, so for every point $$(x,y)$$ on the circle we have

$$ (x-a)^2 + (y-0)^2 = 5^2 = 25. $$

This is the general form of the required circle once $$a$$ is determined.

Step 3 : Impose the condition that the circle passes through (2,3)

Because the point $$(2,3)$$ lies on the circle, its coordinates must satisfy the equation obtained in Step 2:

$$ (2-a)^2 + (3-0)^2 = 25. $$

Simplify:

$$ (2-a)^2 + 9 = 25 \\[2mm] (2-a)^2 = 25 - 9 = 16. $$

Step 4 : Solve for the centre coordinate a

$$ (2-a)^2 = 16 \\[2mm] 2-a = \\pm4. $$

• If $$2-a = 4$$, then $$a = 2-4 = -2.$$
• If $$2-a = -4$$, then $$a = 2+4 = 6.$$

Thus there are two possible centres: $$(-2,0)$$ and $$(6,0).$$

Step 5 : Write the equations of the circles

• For centre $$(-2,0)$$ and radius $$5$$:
  $$ (x - (-2))^2 + (y-0)^2 = 25 \\Rightarrow (x+2)^2 + y^2 = 25. $$
• For centre $$(6,0)$$ and radius $$5$$:
  $$ (x-6)^2 + (y-0)^2 = 25 \\Rightarrow (x-6)^2 + y^2 = 25. $$

Conclusion

The two circles that satisfy all the given conditions are: $$ (x+2)^2 + y^2 = 25 \quad\text{and}\quad (x-6)^2 + y^2 = 25. $$

Step 1 : Write the general equation of a circle

Any circle in the xy–plane can be written as
$$x^2 + y^2 + 2gx + 2fy + c = 0.$$

Step 2 : Use the condition that the circle passes through the origin

Substitute $$x = 0,\;y = 0$$:

$$0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \;\Rightarrow\; c = 0.$$

Step 3 : Use the x-axis intercept

The circle meets the x-axis at the point $$A(a,0)$$, so this point satisfies the equation:

$$a^2 + 0^2 + 2g\,a + 2f\,(0) + 0 = 0 \;\Rightarrow\; a^2 + 2ga = 0.$$

Provided $$a \neq 0,$$ divide by $$2a$$ to obtain

$$g = -\dfrac{a}{2}.$$

Step 4 : Use the y-axis intercept

Similarly, the circle meets the y-axis at $$B(0,b)$$, so

$$0^2 + b^2 + 2g\,(0) + 2f\,b + 0 = 0 \;\Rightarrow\; b^2 + 2fb = 0.$$

Provided $$b \neq 0,$$ divide by $$2b$$ to get

$$f = -\dfrac{b}{2}.$$

Step 5 : Substitute $$g, f, c$$ back into the general equation

$$x^2 + y^2 + 2\Bigl(-\dfrac{a}{2}\Bigr)x + 2\Bigl(-\dfrac{b}{2}\Bigr)y + 0 = 0.$$

Simplify each term:

$$x^2 + y^2 - a x - b y = 0.$$

Therefore, the required equation of the circle is

$$x^2 + y^2 - a x - b y = 0.$$

The general equation of a circle with centre $$C(h,k)$$ and radius $$r$$ is

$$ (x-h)^2 + (y-k)^2 = r^2 $$

Here, the centre is given as $$C(2,2)$$, so $$h=2$$ and $$k=2$$.

First find the radius. The circle passes through the point $$P(4,5)$$, so the radius is the distance $$CP$$.

By the distance formula,

$$ r = \sqrt{(4-2)^2 + (5-2)^2} $$

$$ r = \sqrt{2^2 + 3^2} $$

$$ r = \sqrt{4 + 9} $$

$$ r = \sqrt{13} $$

Hence $$r^2 = 13$$.

Substitute $$h=2$$, $$k=2$$, and $$r^2 = 13$$ into the standard form:

$$ (x-2)^2 + (y-2)^2 = 13 $$

This is the required equation of the circle.

The circle $$x^2 + y^2 = 25$$ has centre $$O(0, 0)$$ and radius $$r = \sqrt{25} = 5$$.

A point lies inside the circle if its distance from the centre is less than $$r$$, on the circle if it equals $$r$$, and outside if it is greater than $$r$$.

Find the distance of $$P(-2.5, 3.5)$$ from the centre:

$$OP = \sqrt{(-2.5 - 0)^2 + (3.5 - 0)^2} = \sqrt{6.25 + 12.25} = \sqrt{18.5}$$

Since $$\sqrt{18.5} \approx 4.3$$, we have $$OP \approx 4.3 < 5 = r$$.

As the distance from the centre is less than the radius, the point lies inside the circle.

The given equation $$y^2 = 8x$$ has $$y$$ squared, so it is a parabola of the standard form $$y^2 = 4ax$$ with vertex at the origin, opening towards the positive $$x$$-direction.

Comparing $$y^2 = 8x$$ with $$y^2 = 4ax$$: $$4a = 8$$, so $$a = 2$$.

For $$y^2 = 4ax$$:

• Focus is $$(a, 0) = (2, 0)$$.
• Axis of the parabola is the $$x$$-axis, i.e. the line $$y = 0$$.
• Directrix is the line $$x = -a$$, i.e. $$x = -2$$.
• Length of the latus rectum is $$4a = 8$$.

Let a variable point on the required parabola be $$P(x, y)$$.

The focus is $$S(2, 0)$$ and the directrix is the vertical line $$x = -2$$.

Step 1 – Write the two distances.

• Distance of $$P$$ from the focus: $$PS = \sqrt{(x - 2)^2 + (y - 0)^2} = \sqrt{(x - 2)^2 + y^2}.$$
• Perpendicular distance of $$P$$ from the straight line $$x = -2$$: $$\text{dist}(P,\;x = -2) = |x + 2|.$$

Step 2 – Apply the definition of a parabola.
For every point on the parabola the two distances are equal, hence

$$\sqrt{(x - 2)^2 + y^2} = |x + 2|.$$ Square both sides to remove the square–root and the absolute value:

$$ (x - 2)^2 + y^2 = (x + 2)^2. $$

Step 3 – Expand and simplify.

Expand both quadratic expressions:

$$ (x^2 - 4x + 4) + y^2 = x^2 + 4x + 4. $$

Cancel $$x^2$$ and the constant $$4$$ occurring on both sides:

$$ -4x + y^2 = 4x. $$

Collect like terms:

$$ y^2 = 8x. $$

Step 4 – Interpretation.
The obtained equation $$y^2 = 8x$$ is in the standard form $$y^2 = 4px$$ with $$4p = 8\;\Rightarrow\;p = 2$$. Thus

• vertex: $$V(0, 0)$$ (mid-point of the focus and directrix),
• axis: the $$x$$-axis, opening to the right,
• distance vertex–focus (parameter) $$p = 2$$, agreeing with the given focus $$S(2, 0)$$.

Therefore, the required parabola is indeed described by the equation $$y^2 = 8x$$.

Step 1 : Recall the definition of a parabola
A parabola is the locus of all points $$P(x, y)$$ that are equidistant from a fixed point called the focus and a fixed straight line called the directrix.

Step 2 : Locate the given data
The vertex is at $$O(0, 0)$$ and the focus is at $$S(0, 2)$$. Because the focus lies on the positive $$y$$-axis and the vertex lies at the origin, the axis of the parabola is the $$y$$-axis.

Step 3 : Find the directrix
The vertex is the mid-point of the segment joining the focus and the point where the axis meets the directrix. Hence the distance from the vertex to the directrix equals the distance from the vertex to the focus.

• Distance from $$O(0, 0)$$ to the focus $$S(0, 2)$$ is $$2$$ units along the positive $$y$$-axis.
• Therefore the directrix must be the horizontal line that lies $$2$$ units below the vertex, i.e. $$y = -2$$.

Step 4 : Apply the definition (locus condition)
For any point $$P(x, y)$$ on the parabola:

(distance $$PS$$ to the focus) = (perpendicular distance from $$P$$ to the line $$y = -2$$)

Write each distance algebraically.

• Distance from $$P(x, y)$$ to the focus $$S(0, 2)$$:

$$PS = \sqrt{(x - 0)^2 + (y - 2)^2}$$

• Perpendicular distance from $$P(x, y)$$ to the line $$y = -2$$:

$$\text{dist} = \lvert y + 2 \rvert$$

Set the two distances equal and square both sides to remove the square root:

$$\sqrt{(x - 0)^2 + (y - 2)^2} = \lvert y + 2 \rvert$$

$$\bigl((x - 0)^2 + (y - 2)^2\bigr) = (y + 2)^2$$

Step 5 : Simplify the equation

$$x^2 + (y - 2)^2 = (y + 2)^2$$

Expand both squares of binomials:

$$x^2 + \bigl(y^2 - 4y + 4\bigr) = y^2 + 4y + 4$$

Cancel $$y^2$$ on both sides and also the constant $$4$$:

$$x^2 - 4y = 4y$$

Add $$4y$$ to both sides:

$$x^2 = 8y$$

Step 6 : Write in standard form
The obtained equation $$x^2 = 8y$$ matches the standard form $$x^2 = 4ay$$, where $$a = 2$$. Indeed the focus is at $$(0, a) = (0, 2)$$ and the directrix is $$y = -a = -2$$, confirming our result.

Therefore, the required equation of the parabola is $$x^2 = 8y$$.

The axis of the required parabola is the $$y$$-axis. For a parabola whose vertex is at the origin and axis is the $$y$$-axis, the standard form is

$$x^2 = 4ay$$

where $$a$$ is the (signed) distance from the vertex to the focus:  • $$a > 0$$ ⇒ the parabola opens upward (y positive).  • $$a < 0$$ ⇒ the parabola opens downward (y negative).

The given point $$(2, -3)$$ lies below the $$x$$-axis, so the parabola must open downward; consequently $$a$$ will turn out negative.

Substitute $$(x, y) = (2, -3)$$ in $$x^2 = 4ay$$:

$$2^2 = 4a(-3)$$

$$4 = -12a$$

$$a = -\dfrac{1}{3}$$

Insert this value of $$a$$ back into the standard equation:

$$x^2 = 4\left(-\dfrac{1}{3}\right)y$$

$$x^2 = -\dfrac{4}{3}y$$

Thus the required parabola is

$$x^2 = -\dfrac{4}{3}y$$

Step 1 : Compare with the standard form

The standard form of a right-opening parabola with vertex at the origin is $$y^2 = 4ax$$.

Given equation: $$y^2 = 12x$$.

On comparison, $$4a = 12 \;\;\Longrightarrow\;\; a = 3$$.

Step 2 : Vertex and axis

• The vertex of the standard parabola $$y^2 = 4ax$$ is the origin, so the vertex here is $$O(0, 0)$$.
• The axis is the line through the vertex and the focus; for the form $$y^2 = 4ax$$ it is the x-axis.
  Hence the axis is $$y = 0$$.

Step 3 : Focus

For $$y^2 = 4ax$$ the focus is $$(a, 0)$$. Here $$a = 3$$, so

Focus $$\Bigl(3,\,0\Bigr)$$.

Step 4 : Directrix

The directrix of $$y^2 = 4ax$$ is the vertical line $$x = -a$$. Therefore

Directrix $$x = -3$$.

Step 5 : Length of the latus rectum

The length of the latus rectum for $$y^2 = 4ax$$ is $$4a$$. Substituting $$a = 3$$ gives

Length of the latus rectum $$4\times 3 = 12$$.

Summary

Step 1 — Compare with the standard form

The standard equation of a parabola that opens upward is $$x^2 = 4ay$$, whose

• vertex is the origin $$O(0,0)$$,
• axis is the y–axis (the line $$x = 0$$),
• focus is $$F(0,a)$$,
• directrix is the line $$y = -a$$,
• length of the latus-rectum is $$4a$$.

Write the given equation in the same shape:

$$x^2 = 6y \implies x^2 = 4ay$$

Hence

$$4a = 6 \;\Longrightarrow\; a = \dfrac{6}{4} = \dfrac{3}{2}.$$

Step 2 — Focus

For $$a = \dfrac{3}{2}$$ the focus is

$$F\bigl(0,\,a\bigr) = \left(0,\,\dfrac{3}{2}\right).$$

Step 3 — Axis

The axis is the y–axis:

$$x = 0.$$

Step 4 — Directrix

The directrix is a horizontal line at a distance $$a$$ below the vertex:

$$y = -a = -\dfrac{3}{2}.$$

Step 5 — Length of the latus rectum

Length = $$4a = 4\times\dfrac{3}{2} = 6.$$

Thus all the required elements have been determined.

The given conic is

$$y^2 = -8x$$

For a parabola that opens to the left, the standard form is

$$y^2 = -4ax$$

where a is a positive constant. By matching the two equations, we get

$$-4a = -8 \;\; \Longrightarrow \;\; 4a = 8 \;\; \Longrightarrow \;\; a = 2$$

Now we use the standard facts for the left-opening parabola $$y^2 = -4ax$$ :

• Focus : $$(-a,\,0)$$
• Axis : the line through the focus and vertex, i.e. the $$x$$-axis $$y=0$$
• Directrix : the vertical line $$x = a$$
• Length of the latus rectum : $$4a$$

Substituting $$a = 2$$ gives

• Focus : $$(-2,\,0)$$
• Axis : $$y = 0$$
• Directrix : $$x = 2$$
• Length of latus rectum : $$4a = 8$$

Given equation: $$x^2 = -16y$$

To identify the elements of the parabola, rewrite it so that it matches one of the four standard forms known at Class 11 level.

For a parabola whose vertex is at the origin and whose axis is the $$y$$-axis, the two possible standard forms are

• $$x^2 = 4ay$$  (opens upward, a > 0)
• $$x^2 = -4ay$$  (opens downward, a > 0)

Comparing the given equation with $$x^2 = -4ay$$, we get

$$-4a = -16 \;\;\Longrightarrow\;\; a = 4$$

(i) Vertex: Because the equation is already in standard form about the origin, the vertex is at $$ (0,0) $$.

(ii) Axis of the parabola: For both forms $$x^2 = \pm 4ay$$ the axis is the line through the vertex parallel to the $$y$$-axis, namely

$$x = 0$$

(iii) Focus: In the form $$x^2 = -4ay$$ the focus is located at $$ (0,-a) $$. Substituting $$a = 4$$ gives

$$\bigl(0,-4\bigr)$$

(iv) Directrix: For $$x^2 = -4ay$$ the directrix is the horizontal line $$y = a$$. Hence

$$y = 4$$

(v) Length of the latus rectum: For every parabola of the form $$x^2 = \pm 4ay$$, the focal distance that defines the latus rectum is $$4a$$. Therefore

$$4a = 4 \times 4 = 16$$

Summary:

• Focus : $$ (0,-4) $$
• Axis : $$x = 0$$ (the $$y$$-axis)
• Directrix : $$y = 4$$
• Length of the latus rectum : $$16$$

Step 1 : Bring the equation to the standard form
For a right–opening parabola the standard form is $$y^{2}=4ax$$, where

• the distance $$a$$ is the focal distance,
• focus  $$F(a,0)$$,
• directrix $$x=-a$$,
• axis the line $$y=0$$ (the x-axis),
• length of the latus-rectum $$4a$$.

Step 2 : Focus
Focus $$F(a,0)=\left(\dfrac{5}{2},0\right).$$

Step 3 : Axis of the parabola
Since the parabola opens to the right, its axis is the x-axis:
$$y=0.$$

Step 4 : Directrix
Directrix $$x=-a \;\Longrightarrow\; x=-\dfrac{5}{2}.$$

Step 5 : Length of the latus rectum
The length is $$4a = 4\left(\dfrac{5}{2}\right)=10.$$

Geometrical description for a diagram: Draw the parabola opening rightward with vertex at the origin, focus at $$\left(\frac52,0\right)$$, directrix the vertical line $$x=-\frac52$$, and mark the chord through the focus perpendicular to the axis as the latus rectum of length 10.

Step 1 : Rewrite the given equation in the standard form

We know the standard form of a parabola whose axis is the y-axis and which opens downward is $$x^2=-4ay$$.

The given equation is $$x^2=-9y$$. Compare this with $$x^2=-4ay$$:

$$-4a=-9 \;\;\Longrightarrow\;\;4a=9 \;\;\Longrightarrow\;\;a=\tfrac{9}{4}.$$

Step 2 : Vertex

For the form $$x^2=-4ay$$ the vertex is at the origin. Hence the vertex is $$(0,0).$$

Step 3 : Focus

For $$x^2=-4ay$$ the focus has coordinates $$(0,-a).$$ Using $$a=\tfrac{9}{4}$$,

$$\text{Focus}=\bigl(0,-\tfrac{9}{4}\bigr).$$

Step 4 : Axis of the parabola

The axis is the line through the vertex and focus. Because the parabola opens along the y-axis, the axis is

$$x=0.$$

Step 5 : Equation of the directrix

For $$x^2=-4ay$$ the directrix is the line $$y=a.$$ Hence

$$\text{Directrix}:\; y=\tfrac{9}{4}.$$

Step 6 : Length of the latus rectum

The length of the latus rectum for $$x^2=-4ay$$ equals $$4a.$$ Therefore

$$\text{Length of latus rectum}=4a=9.$$

Given data

• Focus $$S(6,0)$$
• Directrix $$x=-6$$ (a vertical line)

Step 1 · Choose a general point on the parabola

Let $$P(x, y)$$ be any point that satisfies the definition of the parabola.

Step 2 · Write the two required distances

1. Distance from $$P(x,y)$$ to the focus $$S(6,0)$$:
   $$PS = \sqrt{(x-6)^2 + (y-0)^2} = \sqrt{(x-6)^2 + y^2}$$
2. Perpendicular distance from $$P(x,y)$$ to the directrix $$x=-6$$:
   Since the line is vertical, the perpendicular distance is the horizontal separation, so
   $$PD = |x - (-6)| = |x+6|$$

Step 3 · Impose the parabola definition

For every point on a parabola, the two distances are equal:

Square both sides to remove the square root and absolute value (the sign is immaterial after squaring):

Step 4 · Expand and simplify

Expand both binomials:

Cancel the common terms $$x^2$$ and $$36$$:

Collect like terms:

Step 5 · Interpretation (optional)

• The equation is of the standard form $$y^2 = 4ax$$ with $$4a = 24$$, giving $$a = 6$$.
• Thus the vertex is at $$ (0,0) $$, the axis is the x-axis, the parabola opens to the right and its latus-rectum length is $$4a = 24$$.

Hence, the required parabola is $$y^2 = 24x$$.

Let an arbitrary point on the required parabola be $$P(x, y)$$.

By definition of a parabola,

distance $$(P,\text{ focus}) = \text{distance}(P,\text{ directrix})$$.

The focus is $$F(0,-3)$$ and the directrix is the line $$y = 3$$.

1. Write the two distances.

• Using the distance formula,
  $$PF = \sqrt{(x-0)^2 + (y+3)^2} = \sqrt{x^2 + (y+3)^2}.$$
• The perpendicular distance from $$P(x,y)$$ to the line $$y=3$$ is
  $$\left|y-3\right|.$$

2. Set the distances equal and square both sides.

$$\sqrt{x^2 + (y+3)^2} = |y-3|$$

Squaring eliminates the square root and the absolute value:

$$x^2 + (y+3)^2 = (y-3)^2.$$

3. Expand the squares.

$$(y+3)^2 = y^2 + 6y + 9, \qquad (y-3)^2 = y^2 - 6y + 9.$$

Substituting,

$$x^2 + y^2 + 6y + 9 = y^2 - 6y + 9.$$

4. Simplify.

Cancel $$y^2$$ and $$9$$ from both sides:

$$x^2 + 6y = -6y.$$

Bring like terms together:

$$x^2 + 6y + 6y = 0 \;\;\Longrightarrow\;\; x^2 + 12y = 0.$$

5. Write the parabola in standard form.

$$x^2 = -12y.$$

This is the required equation. (For comparison with the standard form $$x^2 = -4ay$$, observe that $$-4a = -12 \implies a = 3,$$ confirming the given focus at $$(0,-3)$$.)

Given data

Vertex $$V(0,0)$$; focus $$S(3,0)$$.

The line joining the vertex and focus is the $$x$$-axis, hence the parabola opens to the right.

Step 1 – Locate the parameter $$a$$

$$a = VS = \sqrt{(3-0)^2 + (0-0)^2} = 3$$.

Step 2 – Write the standard form

For a right-opening parabola with vertex at the origin, the standard equation is $$y^2 = 4ax$$.

Substituting $$a = 3$$ gives

$$y^2 = 4(3)x = 12x$$.

Step 3 – (Optional) Verify with the focus–directrix definition

The directrix is the line $$x = -a = -3$$.

For any point $$P(x,y)$$ on the curve:

$$\sqrt{(x-3)^2 + y^2} = |x+3|$$.

Squaring:

$$(x-3)^2 + y^2 = (x+3)^2$$

$$x^2 - 6x + 9 + y^2 = x^2 + 6x + 9$$

$$y^2 = 12x$$.

This matches the equation found above.

Therefore, the required parabola is $$y^2 = 12x$$.

Step 1 Identify the axis and the parameter a
Vertex = $$(0,0)$$ and focus = $$(-2,0)$$ lie on the same horizontal line, so the axis of the parabola is the x-axis.
The distance between the vertex and the focus is
$$a = \lvert -2-0 \rvert = 2.$$

Step 2 Choose the standard form
For a parabola whose vertex is at the origin and whose axis is the x-axis, the standard equation is
$$y^2 = 4ax.$$

If the focus is to the right of the vertex, the curve opens to the right and the coefficient of x is positive. Here the focus $$(-2,0)$$ is to the left of the vertex, so the parabola opens to the left. Hence we use
$$y^2 = -4ax.$$

Step 3 Substitute a = 2
$$y^2 = -4(2)x = -8x.$$

Equation of the required parabola
$$y^2 = -8x.$$

Let the vertex be the origin $$ (0,0) $$ and let the axis of the parabola be the $$x$$-axis.

With these specifications the standard forms available are

• $$y^2 = 4 a x$$   (opens to the right)
• $$y^2 = -4 a x$$   (opens to the left)

The curve passes through $$ (2,3) $$. Substitute $$x = 2$$ and $$y = 3$$ in the first form:

$$3^2 = 4 a \cdot 2 \;\; \Longrightarrow \;\; 9 = 8 a \;\; \Longrightarrow \;\; a = \dfrac{9}{8}.$$

Because $$a>0$$, the assumption of a right-opening branch is self-consistent. (If the second form were used we would get $$9 = -8 a$$, implying $$a<0$$, which is not admissible.)

Hence the required equation is

$$y^2 = 4 a x = 4 \left(\dfrac{9}{8}\right) x = \dfrac{9}{2} x.$$

Step 1: Choose the standard form.
Because the parabola is symmetric with respect to the $$y$$-axis and has its vertex at the origin, its axis is the $$y$$-axis. Hence its equation can be taken as $$x^2 = 4ay$$ (opens upward if $$a>0$$, downward if $$a<0$$).

Step 2: Use the given point.
The point $$(5,2)$$ satisfies the equation:
$$5^2 = 4a \cdot 2$$
$$25 = 8a$$
$$a = \frac{25}{8}$$.

Step 3: Obtain the required equation.
Insert $$a = \frac{25}{8}$$ into $$x^2 = 4ay$$:
$$x^2 = 4 \left( \tfrac{25}{8} \right) y$$
$$x^2 = \tfrac{25}{2} y$$.

In explicit $$y$$-form:
$$y = \tfrac{2}{25} x^2$$.

Therefore, the equation of the desired parabola is $$x^2 = \tfrac{25}{2} y$$ (or equivalently $$y = \tfrac{2}{25} x^2$$).

The given conic is
$$\frac{x^2}{25}+\frac{y^2}{9}=1$$

Compare it with the standard form of an ellipse whose major axis is the x-axis:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\;\; a>b>0$$

By inspection

• $$a^2 = 25 \implies a = 5$$
• $$b^2 = 9 \implies b = 3$$

1. Length of the major axis
Major axis = $$2a = 2\times5 = 10$$

2. Length of the minor axis
Minor axis = $$2b = 2\times3 = 6$$

3. Vertices
Because the major axis is on the x-axis, the vertices are obtained by putting $$x=\pm a$$, $$y=0$$:
$$V_1(a,0)=(5,0),\; V_2(-a,0)=(-5,0)$$

4. Foci

The focal distance $$c$$ satisfies $$c^2 = a^2 - b^2$$:

$$c^2 = 25 - 9 = 16 \;\;\Longrightarrow\;\; c = 4$$

Thus the foci are
$$S_1(c,0)=(4,0),\; S_2(-c,0)=(-4,0)$$

5. Eccentricity

$$e = \frac{c}{a} = \frac{4}{5}=0.8$$

6. Length of the latus rectum

The (double) latus-rectum length for an ellipse is $$\dfrac{2b^{2}}{a}$$:

$$\text{Latus rectum} = \frac{2b^2}{a} = \frac{2\times9}{5} = \frac{18}{5}=3.6$$

Hence all requested elements have been determined.

Given ellipse:

$$9x^2 + 4y^2 = 36$$

First write it in the standard form $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ (or the other way round).

1. Divide every term by $$36$$:

   $$\dfrac{9x^2}{36}+\dfrac{4y^2}{36}=1\;\;\Rightarrow\;\;\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$$

2. Identify the denominators:

   • Under $$x^2$$ we have $$4$$, so $$b^2 = 4 \;\Rightarrow\; b = 2$$.
   • Under $$y^2$$ we have $$9$$, so $$a^2 = 9 \;\Rightarrow\; a = 3$$.

   Because $$a^2 > b^2$$ and it is under $$y^2$$, the major axis is vertical (along the $$y$$-axis).

3. Find the distance of each focus from the centre:

   $$c^2 = a^2 - b^2 = 9 - 4 = 5 \;\Rightarrow\; c = \sqrt{5}$$

4. Centre of the ellipse is the origin $$(0,0)$$, so:

   • Vertices (end-points of the major axis): $$(0,\,\pm a) = (0,\,\pm 3)$$
   • Foci: $$(0,\,\pm c) = \bigl(0,\,\pm\sqrt{5}\bigr)$$
   • Length of the major axis: $$2a = 2\times3 = 6$$
   • Length of the minor axis: $$2b = 2\times2 = 4$$
   • Eccentricity: $$e = \dfrac{c}{a} = \dfrac{\sqrt{5}}{3}$$

Thus all required elements of the ellipse are obtained.

The vertices are given as $$(\pm 13,0)$$. For an ellipse, vertices on the major axis have coordinates $$(\pm a,0)$$; hence

$$a = 13 \;\;\Rightarrow\;\; a^{2}=13^{2}=169.$$

The foci are $$(\pm 5,0)$$. For an ellipse centred at the origin with its major axis along the x‑axis the foci are $$(\pm c,0)$$, so

$$c = 5 \;\;\Rightarrow\;\; c^{2}=5^{2}=25.$$

For every ellipse, the semi-minor axis satisfies $$b^{2}=a^{2}-c^{2}.$$ Substituting the known values:

$$b^{2}=169-25=144 \;\;\Rightarrow\;\; b = 12.$$

Because the centre is the origin and the major axis is along the x-axis, the standard form of the equation is

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1.$$

Replacing $$a^{2}$$ and $$b^{2}$$ with $$169$$ and $$144$$ respectively, we obtain the required equation:

$$\frac{x^{2}}{169}+\frac{y^{2}}{144}=1.$$

We are given

• the foci of the ellipse: $$(0,\,\pm 5)$$, which lie on the y-axis,
• the length of the major axis: $$20$$.

Because the foci are on the y-axis, the major axis is vertical and the ellipse is centred at the origin. Hence we use the standard form

$$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,\quad a> b,$$

where

• $$a$$ is the semi-major axis,
• $$b$$ is the semi-minor axis,
• $$c$$ is the distance of each focus from the centre, with the relation $$c^{2}=a^{2}-b^{2}.$$

Step 1 • Find $$a$$ from the major-axis length

The length of the major axis is $$20$$, so

$$2a = 20 \;\Rightarrow\; a = 10.$$

Step 2 • Find $$c$$ from the coordinates of the foci

Each focus is $$5$$ units from the centre: $$c = 5.$$

Step 3 • Determine $$b^{2}$$ using $$c^{2}=a^{2}-b^{2}$$

$$\begin{aligned} c^{2} &= a^{2}-b^{2}\\ 5^{2} &= 10^{2}-b^{2}\\ 25 &= 100-b^{2}\\ b^{2} &= 100-25 = 75. \end{aligned}$$

Step 4 • Write the required equation

Substituting $$a^{2}=100$$ and $$b^{2}=75$$ into the standard form gives

$$\frac{x^{2}}{75}+\frac{y^{2}}{100}=1.$$

Hence, the equation of the ellipse is the above.

Step 1 : Choose the standard form

Since the major axis is along the  $$x$$–axis and no other information about translation is given, we may take the centre of the ellipse to be the origin.
Hence the required ellipse can be written in the standard (horizontal) form

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\; a>b>0.$$

Step 2 : Use the given points

The ellipse passes through the points $$(4,3)$$ and $$(-1,4)$$. Substituting each point gives two equations.

For \[(4,3): \; \displaystyle \frac{4^{2}}{a^{2}}+\frac{3^{2}}{b^{2}}=1 \;\Rightarrow\; \frac{16}{a^{2}}+\frac{9}{b^{2}}=1. \tag{1}

For \](-1,4): \; \displaystyle \frac{(-1)^{2}}{a^{2}}+\frac{4^{2}}{b^{2}}=1 \;\Rightarrow\; \frac{1}{a^{2}}+\frac{16}{b^{2}}=1. \tag{2}

Step 3 : Solve for $$a^{2}$$ and $$b^{2}$$

Introduce the substitutions $$X=\dfrac{1}{a^{2}}$$ and $$Y=\dfrac{1}{b^{2}}$$ to make the algebra clearer. Equations (1) and (2) become

$$16X+9Y=1, \qquad X+16Y=1.$$

From the second equation, $$X=1-16Y.$$ Substitute this in the first:

$$16(1-16Y)+9Y=1 \;\Rightarrow\; 16-256Y+9Y=1 \;\Rightarrow\; 16-247Y=1.$$

$$\Rightarrow\; -247Y=-15 \;\Rightarrow\; Y=\dfrac{15}{247}.$$

Therefore $$\dfrac{1}{b^{2}}=\dfrac{15}{247}\;\Longrightarrow\; b^{2}=\dfrac{247}{15}.$$

Using $$X=1-16Y$$,

$$X=1-16\left(\dfrac{15}{247}\right)=1-\dfrac{240}{247}=\dfrac{7}{247}.$$

Hence $$\dfrac{1}{a^{2}}=\dfrac{7}{247}\;\Longrightarrow\; a^{2}=\dfrac{247}{7}.$$

We do have $$a^{2}>b^{2}$$, so the major axis is indeed along the $$x$$–axis as required.

Step 4 : Write the equation

Insert $$a^{2}=\dfrac{247}{7}$$ and $$b^{2}=\dfrac{247}{15}$$ in the standard form:

$$\frac{x^{2}}{\dfrac{247}{7}}+\frac{y^{2}}{\dfrac{247}{15}}=1.$$

To clear the fractional denominators, multiply every term by $$247$$:

$$7x^{2}+15y^{2}=247.$$

Step 5 : Verification

• For $$(4,3):\;7(4)^{2}+15(3)^{2}=7\times16+15\times9=112+135=247\;\checkmark
• For $$(-1,4):\;7(-1)^{2}+15(4)^{2}=7\times1+15\times16=7+240=247\;\checkmark

Hence the obtained equation satisfies all the given conditions.

For an ellipse in standard form

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\quad (a > b > 0)$$

the important facts are:

• Centre: $$O(0,0)$$
• Vertices on the major axis: $$(\pm a,0)$$    (major axis is the x-axis because $$a>b$$)
• Foci: $$(\pm c,0)$$ where $$c^2=a^2-b^2$$
• Eccentricity: $$e=\dfrac{c}{a}$$
• Length of the major axis: $$2a$$
• Length of the minor axis: $$2b$$
• Length of the latus rectum: $$\dfrac{2b^2}{a}$$

Compare the given equation

$$\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$$

with the standard form:

$$a^2 = 36 \;\Rightarrow\; a = 6, \qquad b^2 = 16 \;\Rightarrow\; b = 4.$$

1. Foci

$$c^2 = a^2 - b^2 = 36 - 16 = 20 \;\Rightarrow\; c = \sqrt{20}=2\sqrt5.$$

Hence the foci are $$(\,\pm 2\sqrt5,\;0\,).$$

2. Vertices

The vertices lie at $$(\pm a,0)=(\pm 6,0).$$

3. Length of the major axis

$$2a = 2\times6 = 12.$$

4. Length of the minor axis

$$2b = 2\times4 = 8.$$

5. Eccentricity

$$e = \dfrac{c}{a} = \dfrac{2\sqrt5}{6} = \dfrac{\sqrt5}{3}.$$

6. Length of the latus rectum

$$\dfrac{2b^2}{a} = \dfrac{2\times16}{6} = \dfrac{32}{6} = \dfrac{16}{3}.$$

All requested elements have now been found.

Compare the given equation with the standard form of an ellipse centred at the origin

$$\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1 \quad (a \ge b > 0).$$

The larger denominator (here 25) is identified with $$a^{2}$$ because the larger semi-axis is called the major one. Hence

$$a^{2} = 25 \;\;\Rightarrow\;\; a = 5,$$

$$b^{2} = 4 \;\;\Rightarrow\;\; b = 2.$$

Since the larger denominator is under $$y^{2}$$, the major axis lies on the y-axis.

1. Vertices

• On the major (y-) axis: $$\bigl(0,\,\pm a\bigr) = (0,\,\pm5).$$
• On the minor (x-) axis (co-vertices, asked in some books): $$\bigl(\pm b,\,0\bigr) = (\pm2,0).$$

2. Foci

The focal distance $$c$$ is obtained from $$c^{2}=a^{2}-b^{2}$$:

$$c^{2}=25-4=21 \;\;\Longrightarrow\;\; c = \sqrt{21}.$$

Because the major axis is vertical, the foci are

$$\bigl(0,\,\pm c\bigr) = \bigl(0,\,\pm\sqrt{21}\bigr).$$

3. Lengths of the axes

• Major axis: $$2a = 2\times5 = 10.$$
• Minor axis: $$2b = 2\times2 = 4.$$

4. Eccentricity

$$e = \dfrac{c}{a} = \dfrac{\sqrt{21}}{5}.$$

5. Length of the latus rectum

For an ellipse, $$\text{latus-rectum length} = \dfrac{2b^{2}}{a}.$$ Hence

$$\dfrac{2b^{2}}{a} = \dfrac{2\times4}{5} = \dfrac{8}{5}.$$

All the required elements have now been determined.

The given equation is

$$ \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1 $$

Compare it with the standard form of an ellipse centred at the origin with major axis along the x-axis:

$$ \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1, \; a^{2}>b^{2} $$

On comparison we obtain

$$ a^{2}=16 \;\Rightarrow\; a=4, \qquad b^{2}=9 \;\Rightarrow\; b=3 $$

1. Eccentricity

$$ e=\sqrt{1-\dfrac{b^{2}}{a^{2}}}=\sqrt{1-\dfrac{9}{16}}=\sqrt{\dfrac{7}{16}}=\dfrac{\sqrt{7}}{4} $$

2. Foci

The distance of each focus from the centre is

$$ c=a e=4\left(\dfrac{\sqrt{7}}{4}\right)=\sqrt{7}. $$

Therefore the foci are

$$ (c,0) \text{ and } (-c,0) \;\Rightarrow\; (\sqrt{7},0),\;(-\sqrt{7},0). $$

3. Vertices

Vertices lie on the major axis at a distance a from the centre:

$$ (a,0),\;(-a,0) \;\Rightarrow\; (4,0),\;(-4,0). $$

4. Length of the major axis

$$ 2a = 2\times4 = 8 $$

5. Length of the minor axis

$$ 2b = 2\times3 = 6 $$

6. Length of the latus rectum

For an ellipse, the length of each latus rectum is

$$ \dfrac{2b^{2}}{a}=\dfrac{2\times9}{4}=\dfrac{9}{2}=4.5 $$

Summary

For a general ellipse with centre at the origin

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \;\;\; (a>b>0)$$

the facts we need are

• Major axis length = $$2a$$; vertices on the major axis are $$(\pm a,0)$$ if it is horizontal and $$(0,\pm a)$$ if it is vertical.
• Minor axis length = $$2b$$.
• Distance of every focus from the centre: $$c$$ where $$c^{2}=a^{2}-b^{2}$$; eccentricity $$e=\dfrac{c}{a}\ (0
• Length of each latus rectum: $$\dfrac{2b^{2}}{a}$$.

The given ellipse is

$$\frac{x^{2}}{25}+\frac{y^{2}}{100}=1.$$

Compare with the standard form:

$$\frac{x^{2}}{25}+\frac{y^{2}}{100}=1\quad\Longrightarrow\quad a^{2}=100,\;b^{2}=25$$

hence

$$a=10,\;b=5.$$

Because the larger denominator is under $$y^{2}$$, the major axis is the y-axis (vertical).

1. Vertices
   On the major (vertical) axis: $$\bigl(0,\pm a\bigr)=(0,\pm 10).$$
2. Foci
   First find $$c$$: $$c^{2}=a^{2}-b^{2}=100-25=75\;\Longrightarrow\;c=\sqrt{75}=5\sqrt{3}.$$
   Thus the foci are $$\bigl(0,\pm c\bigr)=\bigl(0,\pm5\sqrt{3}\bigr).$$
3. Length of the major axis
   $$2a=2\times10=20.$$
4. Length of the minor axis
   $$2b=2\times5=10.$$
5. Eccentricity
   $$e=\frac{c}{a}=\frac{5\sqrt{3}}{10}=\frac{\sqrt{3}}{2}.$$
6. Length of the latus rectum
   $$\frac{2b^{2}}{a}=\frac{2\times25}{10}=5.$$

All required parameters are now determined.

For an ellipse written in the standard form

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\;\;(a>b>0),$$

• the centre is the origin $$O(0,0)$$,
• the transverse (major) axis lies along the $$x$$–axis,
• the semi‑major axis is $$a$$ and the semi‑minor axis is $$b$$,
• $$c^{2}=a^{2}-b^{2}$$,  so foci are $$F_{1}(-c,0),\;F_{2}(c,0)$$,
• eccentricity $$e=\dfrac{c}{a}$$,
• length of the latus-rectum $$\displaystyle \ell=\frac{2b^{2}}{a}$$.

Compare the given ellipse

$$\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$$

with the standard form.

Here

$$a^{2}=49 \;\Rightarrow\; a=7,\qquad b^{2}=36 \;\Rightarrow\; b=6.$$

Step 1: Find c

$$c^{2}=a^{2}-b^{2}=49-36=13\;\;\Rightarrow\;\;c=\sqrt{13}.$$

Step 2: Coordinates of the foci

$$F_{1}\bigl(-\sqrt{13},0\bigr),\;F_{2}\bigl(\sqrt{13},0\bigr).$$

Step 3: Coordinates of the vertices

$$V_{1}(-7,0),\;V_{2}(7,0).$$

Step 4: Lengths of the axes

Major axis: $$2a=2\times7=14.$$
Minor axis: $$2b=2\times6=12.$$

Step 5: Eccentricity

$$e=\frac{c}{a}=\frac{\sqrt{13}}{7}.$$

Step 6: Length of the latus rectum

$$\ell=\frac{2b^{2}}{a}=\frac{2\times36}{7}=\frac{72}{7}.$$

Hence all the required elements have been determined.

We start from the equation of the ellipse

$$\frac{x^2}{100}+\frac{y^2}{400}=1.$$

Compare it with the standard form of an ellipse whose axes are along the coordinate axes and whose major axis is the y-axis:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\;(b\ge a).$$

• From $$a^2=100$$ we get $$a=10.$$
• From $$b^2=400$$ we get $$b=20.$$

1. Vertices

The vertices lie on the major axis (the y-axis):

$$(0,\pm b)=(0,20),\,(0,-20).$$

2. Length of the major axis

$$2b=2\times20=40.$$

3. Length of the minor axis

$$2a=2\times10=20.$$

4. Eccentricity

$$e=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{100}{400}}=\sqrt{\frac34}=\frac{\sqrt3}{2}.$$

5. Foci

The focal distance is $$c=be=20\cdot\frac{\sqrt3}{2}=10\sqrt3.$$

Hence the foci are

$$(0,\pm c)=(0,\pm10\sqrt3).$$

6. Length of the latus rectum

For an ellipse with semi-major axis $$b$$ and semi-minor axis $$a$$,

$$L=\frac{2a^2}{b}=\frac{2\times100}{20}=10.$$

All the required characteristics are thus obtained.

Given ellipse $$36x^2+4y^2=144$$.

1. Reduce to standard form

Divide by $$144$$:

$$\frac{x^2}{4}+\frac{y^2}{36}=1$$

2. Identify the semi-axes

Compare with $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$.

• $$a^2=4 \;\Rightarrow\; a=2$$
• $$b^2=36 \;\Rightarrow\; b=6$$

Since $$b>a$$, the major axis lies on the y-axis, the minor axis on the x-axis and the centre is $$O(0,0)$$.

3. Vertices

• Ends of the major axis (called vertices): $$V_1(0,6),\;V_2(0,-6)$$
• Ends of the minor axis (co-vertices): $$C_1(2,0),\;C_2(-2,0)$$

4. Foci

For an ellipse with major axis $$2b$$:

$$c^2=b^2-a^2\;\Rightarrow\;c^2=36-4=32\;\Rightarrow\;c=4\sqrt2$$

Foci: $$S_1(0,4\sqrt2),\;S_2(0,-4\sqrt2)$$

5. Lengths of the axes

• Major axis length: $$2b=12$$
• Minor axis length: $$2a=4$$

6. Eccentricity

$$e=\frac{c}{b}=\frac{4\sqrt2}{6}=\frac{2\sqrt2}{3}$$

7. Length of the latus rectum

For an ellipse, $$\text{latus rectum}=\frac{2b^2}{c}$$.

$$\frac{2\times36}{4\sqrt2}=\frac{72}{4\sqrt2}=\frac{18}{\sqrt2}=9\sqrt2$$

Summary

We are given the curve

$$16x^2 + y^2 = 16$$

To compare it with the standard form of an ellipse we divide throughout by 16:

$$\frac{x^2}{1} + \frac{y^2}{16} = 1$$

This is of the form

$$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1, \quad a^2 > b^2$$

with centre at the origin. We read off

$$a^2 = 16, \; a = 4, \qquad b^2 = 1, \; b = 1$$

1. Foci
   For an ellipse centred at the origin with the major axis along the y-axis, the foci are \((0,\pm c)\), where $$c^2 = a^2 - b^2$$.
   $$c^2 = 16 - 1 = 15 \;\;\Rightarrow\;\; c = \sqrt{15}$$
   Hence the foci are $$ (0,\;\sqrt{15}) \text{ and } (0,-\sqrt{15}). $$
2. Vertices
   The endpoints of the major axis are at $$ (0,\pm a) = (0,\pm 4). $$
3. Length of the major axis
   $$2a = 2\times 4 = 8.$$
4. Length of the minor axis
   $$2b = 2\times 1 = 2.$$
5. Eccentricity
   $$e = \frac{c}{a} = \frac{\sqrt{15}}{4}.$$
6. Length of the latus rectum
   For an ellipse the length of each latus rectum is $$\frac{2b^2}{a}.$$ Therefore
   $$\text{latus rectum} = \frac{2\times 1}{4} = \frac12.$$

The given curve is $$4x^2 + 9y^2 = 36$$.

Reduce it to the standard form of an ellipse by dividing every term by $$36$$:

$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$

Comparing with the standard equation $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\;(a^2 \gt b^2)$$ we get

• $$a^2 = 9 \;\Rightarrow\; a = 3$$
• $$b^2 = 4 \;\Rightarrow\; b = 2$$

The ellipse is centred at the origin and its major axis lies along the $$x$$-axis (because $$a^2>b^2$$).

Vertices (end-points of the major axis)
$$\bigl(\pm a,0\bigr) = (\pm 3,0)$$

Length of the major axis
$$2a = 2\times3 = 6$$

Length of the minor axis
$$2b = 2\times2 = 4$$

Distance of each focus from the centre
$$c^2 = a^2 - b^2 = 9 - 4 = 5$$ hence $$c = \sqrt5.$$

Coordinates of the foci
$$\bigl(\pm c,0\bigr) = (\pm\sqrt5,0)$$

Eccentricity
$$e = \frac{c}{a} = \frac{\sqrt5}{3}$$

Length of the latus rectum
$$\text{L.R.} = \frac{2b^2}{a} = \frac{2\times4}{3} = \frac{8}{3}$$

The ellipse has its two vertices at $$(\pm 5,0)$$ and the two foci at $$(\pm 4,0)$$, which are all situated on the x-axis and are symmetric about the origin. Therefore the centre of the ellipse is the origin $$(0,0)$$ and the major axis lies along the x-axis.

For an ellipse whose centre is at the origin and whose major axis is the x-axis, the standard form of the equation is

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$$

where

• $$a$$ is the semi‑major axis (distance from the centre to each vertex),
• $$b$$ is the semi‑minor axis,
• $$c$$ is the focal distance (distance from the centre to each focus), and
• the fundamental relation $$c^{2}=a^{2}-b^{2}$$ always holds for an ellipse.

Step 1 — Find $$a$$.
The vertices are $$(\pm 5,0)$$, so $$a=5$$.

Step 2 — Find $$c$$.
The foci are $$(\pm 4,0)$$, so $$c=4$$.

Step 3 — Compute $$b^{2}$$ using $$c^{2}=a^{2}-b^{2}$$.

\[\begin{aligned} c^{2} &= a^{2}-b^{2} \\ 4^{2} &= 5^{2}-b^{2} \\ 16 &= 25-b^{2} \\ b^{2} &= 25-16 = 9 \end{aligned}\]

Hence $$b=3$$ (positive length).

Step 4 — Write the required equation.

$$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1.$$

This is the only ellipse that satisfies the given vertices and foci.

Step 1 : Locate the centre and the major axis
The vertices are $$(0,\,\pm 13)$$ and the foci are $$(0,\,\pm 5)$$.
Both pairs lie on the y-axis, so the centre is their midpoint, namely $$(0,0)$$, and the major axis is the y-axis.

Step 2 : Identify the semi-major and focal distances
For an ellipse, the vertices are one semi-major length away from the centre, while the foci are one focal length away.

• Semi-major length: $$a = 13$$ (distance from $$(0,0)$$ to $$(0,\pm 13)$$)
• Focal length: $$c = 5$$ (distance from $$(0,0)$$ to $$(0,\pm 5)$$)

Step 3 : Compute the semi-minor length
For every ellipse, $$c^2 = a^2 - b^2$$, so

$$b^2 = a^2 - c^2 = 13^2 - 5^2 = 169 - 25 = 144\;\Rightarrow\; b = 12.$$

Step 4 : Write the standard equation
Because the major axis is vertical (along the y-axis) and the centre is $$(0,0)$$, the standard form is

$$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1.$$

Substituting $$a^2 = 169$$ and $$b^2 = 144$$ gives

$$\boxed{\displaystyle \frac{x^2}{144} + \frac{y^2}{169} = 1}. $$

This ellipse has the required vertices and foci.

For a standard ellipse with its centre at the origin and its major axis along the x-axis we use

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a> b>0,$$

where

• $$\pm(a,0)$$ are the vertices on the major axis,
• $$\pm(c,0)$$ are the foci, with $$c^2=a^2-b^2$$.

From the data

Vertices: $$(\pm6,0)\;\Rightarrow\;a=6\;\Rightarrow\;a^2=36.$$

Foci: $$(\pm4,0)\;\Rightarrow\;c=4\;\Rightarrow\;c^2=16.$$

Compute $$b^2$$ using $$c^2=a^2-b^2$$:

$$b^2=a^2-c^2=36-16=20.$$

Substituting $$a^2=36$$ and $$b^2=20$$ in the standard form, we obtain the required equation:

$$\frac{x^2}{36}+\frac{y^2}{20}=1.$$

Given data

• Ends of the major axis: $$(3,0)$$ and $$(-3,0)$$
• Ends of the minor axis: $$(0,2)$$ and $$(0,-2)$$

1. Locate the centre

The centre of an ellipse is the midpoint of the ends of either axis.

Midpoint of the major-axis ends:

$$\Big(\tfrac{3+(-3)}{2},\;\tfrac{0+0}{2}\Big)=(0,0)$$

Hence the centre is the origin $$O(0,0)$$.

2. Identify the directions of the axes

• The major axis joins $$(-3,0)\to(3,0)$$. It therefore lies on the $$x$$-axis.
• The minor axis joins $$(0,-2)\to(0,2)$$. It therefore lies on the $$y$$-axis.

3. Read off the semi-lengths

• Semi-major axis length: $$a=3$$ (distance from the centre to either end of the major axis).
• Semi-minor axis length: $$b=2$$ (distance from the centre to either end of the minor axis).

4. Standard form about the origin

For an ellipse centred at the origin with the major axis along the $$x$$-axis, the standard equation is

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$ Substituting $$a=3$$ and $$b=2$$ gives

$$\frac{x^2}{3^2}+\frac{y^2}{2^2}=1.$$ Simplify:

$$\frac{x^2}{9}+\frac{y^2}{4}=1.$$

5. Verification (optional)

Check that the given end–points satisfy the equation:

• Point $$(3,0)$$: $$\tfrac{3^2}{9}+\tfrac{0^2}{4}=1+0=1.$$
• Point $$(0,2)$$: $$\tfrac{0^2}{9}+\tfrac{2^2}{4}=0+1=1.$$

Both satisfy the equation, confirming correctness.

Therefore, the required ellipse is

$$\frac{x^2}{9}+\frac{y^2}{4}=1.$$

We are told

• the ends of the major axis are $$(0,\,\pm\sqrt{5})$$,
• the ends of the minor axis are $$(\pm1,\,0)$$.

Both pairs are symmetric about the origin, so the centre of the ellipse is the origin $$O(0,0)$$.

Step 1 – Semi-axes lengths

Distance from the centre to either end of the major axis:

$$a=\sqrt{5}\;\Rightarrow\;a^{2}=5$$

Distance from the centre to either end of the minor axis:

$$b=1\;\Rightarrow\;b^{2}=1$$

Step 2 – Standard form

The major axis is vertical (along the y-axis), so the standard form of an ellipse centred at the origin is

$$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$$

Substituting $$a^{2}=5$$ and $$b^{2}=1$$ gives

$$\frac{x^{2}}{1}+\frac{y^{2}}{5}=1$$

Step 3 – Final equation

Simplifying, the required equation is

$$x^{2}+\frac{y^{2}}{5}=1$$

For any ellipse with centre at the origin and major axis along the $$x$$-axis, we use the standard form

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a>b>0$$

where

• $$2a$$ is the length of the major axis,
• the foci are $$(\pm c,0)$$ with $$c^{2}=a^{2}-b^{2}$$.

Step 1 · Find $$a$$ from the major axis

Given length of major axis = $$26$$, therefore

$$2a=26 \;\Rightarrow\; a=\frac{26}{2}=13$$

Step 2 · Read $$c$$ from the foci

The foci are $$(\pm5,0)$$, so

$$c=5$$

Step 3 · Compute $$b$$ using $$c^{2}=a^{2}-b^{2}$$

Substituting $$a=13$$ and $$c=5$$:

$$25=a^{2}-b^{2}=169-b^{2}$$

$$b^{2}=169-25=144\quad\Rightarrow\quad b=12$$ (positive length)

Step 4 · Write the required equation

Insert $$a^{2}=169$$ and $$b^{2}=144$$ in the standard form:

$$\frac{x^{2}}{169}+\frac{y^{2}}{144}=1$$

This ellipse has centre at the origin, major axis $$26$$ along the $$x$$-axis, and foci $$(\pm5,0)$$, satisfying all the given conditions.

Step 1 : Locate the centre and the focal distance
The foci are given as $$(0,\,\pm 6)$$, which are equidistant from the origin on the y-axis. Hence the centre of the ellipse is the origin $$O(0,0)$$ and the semi–focal distance is $$c=6$$.

Step 2 : Use the length of the minor axis
The length of the minor axis is $$16$$, so the semi-minor axis is
$$b=\frac{16}{2}=8\;\Longrightarrow\;b^{2}=64$$.

Step 3 : Find the semi-major axis
For an ellipse centred at the origin with its major axis along the y-axis, the fundamental relation is
$$c^{2}=a^{2}-b^{2}$$.
Substituting $$c=6$$ and $$b^{2}=64$$ gives
$$36=a^{2}-64\;\Longrightarrow\;a^{2}=36+64=100\;\Longrightarrow\;a=10\;(a>b)$$.

Step 4 : Write the required equation
Because the major axis is vertical, the standard form is
$$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$$.
Replacing $$a^{2}$$ and $$b^{2}$$ by their values, we get
$$\frac{x^{2}}{64}+\frac{y^{2}}{100}=1$$.

Thus, the equation of the ellipse satisfying the given conditions is obtained.

Given data

• Foci: $$(\pm 3,0)$$
• Semi-major axis length: $$a = 4$$

1. Locate the centre and the major axis
Because the foci are symmetric with respect to the origin, the midpoint of $$(-3,0)$$ and $$(3,0)$$ is $$(0,0)$$. Hence the centre is at the origin.
Both foci lie on the x-axis, so the major axis is the x-axis.

2. Identify the linear eccentricity
The distance from the centre to either focus is $$c = 3$$.

3. Use the relation $$c^{2} = a^{2} - b^{2}$$
Substituting the known values:

$$3^{2} = 4^{2} - b^{2} \; \Longrightarrow \; 9 = 16 - b^{2} \; \Longrightarrow \; b^{2} = 7$$

4. Write the standard form
For an ellipse centred at the origin with its major axis along the x-axis: $$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$$.

Putting $$a^{2} = 16$$ and $$b^{2} = 7$$ we get
$$\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$$.

This is the required equation of the ellipse.

Step 1 – Recall the basic relation
For an ellipse with semi-major axis $$a$$, semi-minor axis $$b$$ and linear eccentricity $$c$$ (distance from centre to either focus), the parameters satisfy
$$c^{2}=a^{2}-b^{2}$$

Step 2 – Insert the given data
The question gives $$b=3$$ and $$c=4$$, so
$$c^{2}=4^{2}=16,\; b^{2}=3^{2}=9$$

Step 3 – Compute $$a^{2}$$
Substitute into $$c^{2}=a^{2}-b^{2}$$:
$$16=a^{2}-9\;\;\Longrightarrow\;\; a^{2}=16+9=25$$
Since $$a^{2}>b^{2}$$ (indeed $$25>9$$), the value is admissible.

Step 4 – Choose the correct standard form
Because the foci lie on the $$x$$-axis, the major axis is the $$x$$-axis. With centre at the origin, the standard equation is
$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$

Step 5 – Write the required equation
Insert $$a^{2}=25,\; b^{2}=9$$:
$$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$$

The ellipse whose centre is at the origin, foci at $$(\pm4,0)$$ and semi-minor axis $$3$$ is therefore represented by the equation above.

Let the required ellipse have semi‑major axis length $$a$$ and semi‑minor axis length $$b$$.

Because the centre is at the origin and the major axis lies on the $$y$$-axis, its standard equation is

$$\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1,\qquad a > b > 0.$$

Both given points must satisfy this equation.

For $$(3, 2)$$:

$$\frac{3^{2}}{b^{2}} + \frac{2^{2}}{a^{2}} = 1 \;\Longrightarrow\; \frac{9}{b^{2}} + \frac{4}{a^{2}} = 1 \quad (1)$$

For $$(1, 6)$$:

$$\frac{1^{2}}{b^{2}} + \frac{6^{2}}{a^{2}} = 1 \;\Longrightarrow\; \frac{1}{b^{2}} + \frac{36}{a^{2}} = 1 \quad (2)$$

Introduce the abbreviations $$A = a^{2},\; B = b^{2}.$$ Equations (1) and (2) become

$$\frac{9}{B} + \frac{4}{A} = 1 \quad (1')$$ $$\frac{1}{B} + \frac{36}{A} = 1 \quad (2').$$

Step 1 – Eliminate $$1/B$$.

From (1′): $$\dfrac{1}{B} = \dfrac{1-\dfrac{4}{A}}{9}.$$

Set this equal to the $$1/B$$ obtained from (2′):

$$\dfrac{1-\dfrac{4}{A}}{9} = 1 - \dfrac{36}{A}.$$

Multiply by 9 to clear the denominator:

$$1 - \frac{4}{A} = 9 - \frac{324}{A}.$$

Bring all terms to one side:

$$1 - \frac{4}{A} - 9 + \frac{324}{A} = 0 \;\Longrightarrow\; -8 + \frac{320}{A} = 0.$$

Hence

$$\frac{320}{A} = 8 \;\Longrightarrow\; A = 40.$$

Thus $$a^{2} = 40.$

Step 2 – Find $$b^{2}$$.

Substitute $$A = 40$$ in (2′):

$$\frac{1}{B} + \frac{36}{40} = 1 \;\Longrightarrow\; \frac{1}{B} + 0.9 = 1.$$

Therefore

$$\frac{1}{B} = 0.1 = \frac{1}{10} \;\Longrightarrow\; B = 10.$$

So $$b^{2} = 10.$$ Since $$a^{2} = 40 > 10 = b^{2},$$ the major axis is indeed along the $$y$$-axis as required.

Step 3 – Equation of the ellipse.

Insert $$a^{2}=40,\;b^{2}=10$$ in the standard form:

$$\frac{x^{2}}{10} + \frac{y^{2}}{40} = 1.$$

Multiplying by 40 if desired gives the equivalent form $$4x^{2} + y^{2} = 40.$

Hence, the ellipse sought is

$$\frac{x^{2}}{10} + \frac{y^{2}}{40} = 1.$$

Given The ellipse has its major axis on the $$x$$-axis and it passes through the points $$(4,3)$$ and $$(6,2)$$.

With the centre at the origin and the major axis along the $$x$$-axis, the standard equation is

$$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, \; a > b > 0$$

Because the two given points lie on the ellipse, each must satisfy the equation.

1. Substituting $$(4,3)$$:

$$\frac{4^{2}}{a^{2}} + \frac{3^{2}}{b^{2}} = 1 \;\Rightarrow\; \frac{16}{a^{2}} + \frac{9}{b^{2}} = 1$$

2. Substituting $$(6,2)$$:

$$\frac{6^{2}}{a^{2}} + \frac{2^{2}}{b^{2}} = 1 \;\Rightarrow\; \frac{36}{a^{2}} + \frac{4}{b^{2}} = 1$$

Introduce $$m = \dfrac{1}{a^{2}}$$ and $$n = \dfrac{1}{b^{2}}$$ to obtain a pair of linear equations:

$$16m + 9n = 1 \;\; \text{(i)}$$

$$36m + 4n = 1 \;\; \text{(ii)}$$

Solving (i) and (ii)

• Multiply (i) by 4: $$64m + 36n = 4$$
• Multiply (ii) by 9: $$324m + 36n = 9$$

Subtract the first new equation from the second:

$$(324m - 64m) = 9 - 4 \;\Rightarrow\; 260m = 5$$

$$m = \frac{5}{260} = \frac{1}{52}$$

Therefore $$a^{2} = \frac{1}{m} = 52$$.

Substitute $$m = \tfrac{1}{52}$$ into equation (i):

$$16\left(\frac{1}{52}\right) + 9n = 1$$

$$\frac{16}{52} + 9n = 1 \;\Rightarrow\; \frac{4}{13} + 9n = 1$$

$$9n = 1 - \frac{4}{13} = \frac{9}{13} \;\Rightarrow\; n = \frac{1}{13}$$

Hence $$b^{2} = \frac{1}{n} = 13$$.

Because $$a^{2} = 52 > b^{2} = 13$$, the major axis is indeed along the $$x$$-axis, consistent with the condition.

Equation of the ellipse

$$\frac{x^{2}}{52} + \frac{y^{2}}{13} = 1$$

Step 1  Identify the orientation and centre
The foci are $$(0,\,\pm 3)$$ and the vertices are $$(0,\,\pm\dfrac{\sqrt{11}}{2})$$; both lie on the y-axis and are symmetric about the origin. Hence the centre of the hyperbola is $$O(0,0)$$ and the transverse axis is vertical.

Step 2  Write the standard form
For a hyperbola with centre at the origin and transverse axis along the y-axis, the standard equation is
$$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1,$$
where

• $$(0,\,\pm a)$$ are the vertices,
• $$(0,\,\pm c)$$ are the foci,
• and $$c^{2}=a^{2}+b^{2}.$$

The two foci lie on the y–axis at $$F_1(0,12)$$ and $$F_2(0,-12)$$, so the centre is the origin and the transverse axis is the y–axis.

With this orientation the standard form of the hyperbola is

$$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\;,$$

where

• $$a$$ = semi-transverse axis length, so the vertices are $$V_1(0,a)$$ and $$V_2(0,-a)$$;
• $$c$$ = distance of each focus from the centre, related by $$c^2=a^2+b^2$$.

1. Using the foci

The given foci give $$c=12$$, hence

\[c^2=12^2=144 \;\;\Longrightarrow\;\; a^2+b^2=144 \tag{1}\]

2. Using the length of the latus rectum

For any hyperbola $$\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$$, the length of each latus rectum is

$$\text{LR}=\frac{2b^2}{a}.$$

Given $$\text{LR}=36$$, so

\[\frac{2b^2}{a}=36 \;\;\Longrightarrow\;\; b^2=18a. \tag{2}\]

3. Solving for a and b

Substitute $$b^2=18a$$ from (2) into (1):

$$a^2+18a=144$$

$$a^2+18a-144=0$$

$$a=\frac{-18\pm\sqrt{18^2+4\cdot144}}{2}=\frac{-18\pm30}{2}.$$

The positive root is required (a is a length):

$$a=6 \;\;\Longrightarrow\;\; a^2=36.$$

Now, from (2),

$$b^2=18a=18\times6=108.$$

4. Equation of the hyperbola

Substituting $$a^2=36$$ and $$b^2=108$$ in the standard form gives

$$\frac{y^2}{36}-\frac{x^2}{108}=1.$$

One may also write it as $$3y^2 - x^2 = 108$$, but the first is the recognised standard form.

The given equation is

$$\frac{x^2}{16}-\frac{y^2}{9}=1.$$

Compare it with the standard form of a rectangular hyperbola whose transverse axis is the x-axis,

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$$

Hence

$$a^2 = 16 \;\;\Rightarrow\;\; a = 4,$$

$$b^2 = 9 \;\;\Rightarrow\;\; b = 3.$$

1. Vertices

The vertices lie at $$ (\pm a,0). $$ Therefore

• First vertex : $$(4,0)$$
• Second vertex : $$(-4,0)$$

2. Foci

For a hyperbola in this form, $$c^2 = a^2 + b^2.$$

So

$$c^2 = 16 + 9 = 25 \;\;\Longrightarrow\;\; c = 5.$$

The foci are situated at $$(\pm c,0).$$ Hence

• First focus : $$(5,0)$$
• Second focus : $$(-5,0)$$

3. Eccentricity

The eccentricity of a hyperbola is $$e = \frac{c}{a}.$$ Therefore

$$e = \frac{5}{4}=1.25.$$

4. Length of the latus rectum

For a hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$ the length of the latus rectum is

$$\text{latus rectum}=\frac{2b^2}{a}.$$

Substituting the values,

$$\text{latus rectum}=\frac{2\times 9}{4}=\frac{18}{4}=\frac{9}{2}=4.5.$$

Final results

• Vertices : $$(4,0),\;(-4,0)$$
• Foci : $$(5,0),\;(-5,0)$$
• Eccentricity : $$e = \dfrac{5}{4}$$
• Length of latus rectum : $$\dfrac{9}{2}$$

The given hyperbola is $$\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$$.

Compare it with the standard vertical–transverse form $$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$$, which has its centre at the origin, vertices on the y-axis at $$(0,\pm a)$$, foci at $$(0,\pm c)$$ where $$c^{2}=a^{2}+b^{2}$$, eccentricity $$e=\dfrac{c}{a}$$ and length of the latus rectum $$\dfrac{2b^{2}}{a}$$.

Identify the parameters:

• From $$\frac{y^{2}}{9}$$ we get $$a^{2}=9 \;\Rightarrow\; a=3$$.
• From $$\frac{x^{2}}{27}$$ we get $$b^{2}=27 \;\Rightarrow\; b=\sqrt{27}=3\sqrt{3}$$.

Compute $$c$$:

$$c^{2}=a^{2}+b^{2}=9+27=36 \;\Rightarrow\; c=6.$$

Eccentricity:

$$e=\frac{c}{a}=\frac{6}{3}=2.$$

Vertices (on the y-axis):

$$(0,\,\pm a)=(0,\,3) \text{ and } (0,\,-3).$$

Foci (also on the y-axis):

$$(0,\,\pm c)=(0,\,6) \text{ and } (0,\,-6).$$

Length of the latus rectum:

$$\frac{2b^{2}}{a}=\frac{2\times27}{3}=\frac{54}{3}=18.$$

Hence, all required elements are obtained.

Given equation: $$9y^2 - 4x^2 = 36$$

1. Bring it to standard form

Divide both sides by $$36$$:

$$\frac{9y^2}{36} - \frac{4x^2}{36} = 1 \;\;\Longrightarrow\;\; \frac{y^2}{4} - \frac{x^2}{9} = 1$$

Comparing with the standard equation of a hyperbola whose transverse axis is along the y-axis,

$$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1,$$

we have

• $$a^2 = 4 \;\Longrightarrow\; a = 2,$$
• $$b^2 = 9 \;\Longrightarrow\; b = 3.$$

2. Find c (distance of a focus from the centre)

For a hyperbola, $$c^2 = a^2 + b^2.$$ Hence

$$c^2 = 4 + 9 = 13 \;\Longrightarrow\; c = \sqrt{13}.$$

3. Coordinates of the foci

Because the transverse axis is vertical, the foci are $$\bigl(0,\, \pm c\bigr).$$ Therefore

Foci  : $$\bigl(0,\, +\sqrt{13}\bigr)$$ and $$\bigl(0,\, -\sqrt{13}\bigr).$$

4. Coordinates of the vertices

Vertices lie at $$\bigl(0,\, \pm a\bigr).$$ Hence

Vertices : $$\bigl(0,\, +2\bigr)$$ and $$\bigl(0,\, -2\bigr).$$

5. Eccentricity

$$e = \frac{c}{a} = \frac{\sqrt{13}}{2}.$$

6. Length of the latus rectum

For a hyperbola, the length of each latus rectum is $$\displaystyle \ell = \frac{2b^2}{a}.$$ Thus

$$\ell = \frac{2\times 9}{2} = 9.$$

Summary

• Vertices : $$\bigl(0,\, \pm 2\bigr)$$
• Foci : $$\bigl(0,\, \pm \sqrt{13}\bigr)$$
• Eccentricity : $$e = \dfrac{\sqrt{13}}{2}$$
• Length of latus rectum : $$9$$

Given hyperbola:

$$16x^2-9y^2=576$$

Step 1 ∙ Reduce to standard form

Divide by $$576$$ on both sides:

$$\frac{16x^2}{576}-\frac{9y^2}{576}=1$$

Simplify each fraction:

$$\frac{x^2}{36}-\frac{y^2}{64}=1$$

Hence the equation is in the standard form $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$ with

• $$a^2=36 \implies a=6$$
• $$b^2=64 \implies b=8$$

Step 2 ∙ Find c (focal distance) and eccentricity

For a hyperbola of this form, $$c^2=a^2+b^2$$. Therefore

$$c^2=36+64=100 \;\;\Rightarrow\;\; c=10$$

Eccentricity $$e=\dfrac{c}{a}=\dfrac{10}{6}=\dfrac{5}{3}$$.

Step 3 ∙ Vertices and foci

Because the transverse axis is the $$x$$-axis:

• Vertices: $$(\pm a,0)=(6,0),\;(-6,0)$$
• Foci: $$(\pm c,0)=(10,0),\;(-10,0)$$

Step 4 ∙ Length of the latus rectum

Length $$=\dfrac{2b^2}{a}=\dfrac{2\times64}{6}=\dfrac{128}{6}=\dfrac{64}{3}$$.

Summary

Given equation:
$$5y^2-9x^2=36$$

1. Convert to standard form

Divide by $$36$$: $$\frac{5y^2}{36}-\frac{9x^2}{36}=1\;\Longrightarrow\;\frac{y^2}{\dfrac{36}{5}}-\frac{x^2}{4}=1$$ Thus the hyperbola is of the form $$\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$$ with

• $$a^2=\dfrac{36}{5}\;\;\Rightarrow\;\;a=\dfrac{6}{\sqrt5}$$
• $$b^2=4\;\;\Rightarrow\;\;b=2$$

2. Coordinates of the vertices

For $$y^2/a^2-x^2/b^2=1$$ the transverse axis is the y-axis, so $$\text{Vertices:}\;(0,\pm a)=\left(0,\pm\dfrac{6}{\sqrt5}\right)$$

3. Coordinates of the foci

For a hyperbola, $$c^2=a^2+b^2$$. Hence

$$c^2=\dfrac{36}{5}+4=\dfrac{36}{5}+\dfrac{20}{5}=\dfrac{56}{5}$$ $$\Longrightarrow\;c=\sqrt{\dfrac{56}{5}}=\dfrac{\sqrt{56}}{\sqrt5}=\dfrac{\sqrt{280}}{5}$$

Therefore $$\text{Foci:}\;(0,\pm c)=\left(0,\pm\dfrac{\sqrt{56}}{\sqrt5}\right)=\left(0,\pm\dfrac{\sqrt{280}}{5}\right)$$

4. Eccentricity

$$e=\dfrac{c}{a}=\dfrac{\sqrt{56}/\sqrt5}{6/\sqrt5}=\dfrac{\sqrt{56}}{6}=\dfrac{2\sqrt{14}}{6}=\dfrac{\sqrt{14}}{3}$$

5. Length of the latus rectum

For this orientation, length $$=\dfrac{2b^2}{a}$$:
$$\text{L.R.}=\dfrac{2\cdot4}{6/\sqrt5}=\dfrac{8\sqrt5}{6}=\dfrac{4\sqrt5}{3}$$

Summary

Given hyperbola: $$49y^2-16x^2=784$$.

1. Reduce to the standard form

Divide every term by 784:

$$\frac{49y^2}{784}-\frac{16x^2}{784}=1$$

Simplify both fractions:

$$\frac{y^2}{16}-\frac{x^2}{49}=1$$

This matches the standard vertical hyperbola

$$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$$

Thus

$$a^2=16\;\Longrightarrow\;a=4,\qquad b^2=49\;\Longrightarrow\;b=7$$

2. Vertices

For the above form the vertices lie on the y-axis at $$(0,\pm a)$$.

Hence the vertices are $$(0,4)\;\text{and}\;(0,-4)$$.

3. Eccentricity

$$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{49}{16}}=\sqrt{\frac{65}{16}}=\frac{\sqrt{65}}{4}$$

4. Foci

For this orientation the foci are at $$(0,\pm ae)$$.

Calculate $$ae=4\left(\frac{\sqrt{65}}{4}\right)=\sqrt{65}$$, so

foci: $$(0,\sqrt{65})\;\text{and}\;(0,-\sqrt{65})$$.

5. Length of the latus rectum

Length $$=\dfrac{2b^2}{a}=\dfrac{2\cdot49}{4}=\dfrac{98}{4}=\dfrac{49}{2}$$.

Summary

• Vertices: $$(0,\pm4)$$
• Foci: $$(0,\pm\sqrt{65})$$
• Eccentricity: $$e=\dfrac{\sqrt{65}}{4}$$
• Length of latus rectum: $$\dfrac{49}{2}$$

The vertices are given as $$(\pm 2,0)$$. Their mid-point is the origin, so the centre of the hyperbola is $$O(0,0)$$ and the transverse axis lies on the x-axis.

For a hyperbola with this orientation we start with the standard form

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$.

Step 1: Find $$a$$ from the vertices
The distance from the centre to each vertex equals $$a$$.
Therefore $$a = 2 \;\Rightarrow\; a^2 = 4$$.

Step 2: Find $$c$$ from the foci
The foci are $$(\pm 3,0)$$, so the distance from the centre to either focus is
$$c = 3 \;\Rightarrow\; c^2 = 9$$.

Step 3: Calculate $$b^2$$
For every hyperbola $$c^2 = a^2 + b^2$$. Hence

$$b^2 = c^2 - a^2 = 9 - 4 = 5$$.

Step 4: Write the required equation
Insert $$a^2 = 4$$ and $$b^2 = 5$$ in the standard form:

$$\frac{x^2}{4}-\frac{y^2}{5}=1$$.

Step 1 : Identify the type and the centre
The vertices and foci are on the y-axis and occur in opposite pairs, so the transverse axis is vertical and the centre is the origin $$O(0,0)$$. Hence the required curve is a hyperbola whose standard form is

\[\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\tag{1}\]

Step 2 : Use the vertices to find $$a$$
For the hyperbola (1) the vertices are $$\bigl(0,\pm a\bigr)$$. The given vertices are $$\bigl(0,\pm5\bigr)$$, so

\[a = 5\;\Longrightarrow\;a^{2}=5^{2}=25.\tag{2}\]

Step 3 : Use the foci to find $$c$$
For (1) the foci are $$\bigl(0,\pm c\bigr)$$. The given foci are $$\bigl(0,\pm8\bigr)$$, so

\[c = 8\;\Longrightarrow\;c^{2}=8^{2}=64.\tag{3}\]

Step 4 : Relate $$a,\,b,\,c$$
For every hyperbola,

\[c^{2}=a^{2}+b^{2}.\tag{4}\]

Substituting (2) and (3) into (4):

\[64 = 25 + b^{2}\;\Longrightarrow\;b^{2}=64-25=39.\tag{5}\]

Step 5 : Write the required equation
Insert $$a^{2}=25$$ and $$b^{2}=39$$ in (1):

\[\frac{y^{2}}{25}-\frac{x^{2}}{39}=1.\tag{6}\]

Equation (6) satisfies all the given conditions, so it is the required hyperbola.

Given data

• Vertices : $$(0,\,\pm 3)$$
• Foci    : $$(0,\,\pm 5)$$

1. Locate the centre and the transverse axis

The two vertices are symmetric about the origin and lie on the y-axis; hence

• Centre $$C$$ is the origin $$(0,0)$$.
• The transverse axis is the y-axis.

Therefore the required hyperbola has the standard form

$$\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$$  (axis along y, centre at origin).

2. Use the vertices to obtain $$a$$

For this orientation, the vertices are $$(0,\,\pm a)$$. Comparing with $$(0,\,\pm 3)$$ gives

$$a = 3 \;\;\Rightarrow\;\; a^{2} = 9.$$

3. Use the foci to obtain $$c$$

For the same orientation, the foci are $$(0,\,\pm c)$$. Comparing with $$(0,\,\pm 5)$$ gives

$$c = 5 \;\;\Rightarrow\;\; c^{2} = 25.$$

4. Find $$b^{2}$$ from the relation $$c^{2} = a^{2} + b^{2}$$

Substitute the known values:

$$25 = 9 + b^{2} \;\;\Longrightarrow\;\; b^{2} = 25 - 9 = 16.$$

5. Write the required equation

Insert $$a^{2} = 9$$ and $$b^{2} = 16$$ in the standard form:

$$\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1.$$

Hence the equation of the hyperbola that has vertices $$(0,\,\pm 3)$$ and foci $$(0,\,\pm 5)$$ is

$$\boxed{\dfrac{y^{2}}{9} - \dfrac{x^{2}}{16} = 1}.$$

For a hyperbola centred at the origin whose transverse axis lies on the X-axis, the standard form is

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$$

where

• the foci are $$\bigl(\pm c,\,0\bigr)$$, so $$c$$ is the distance of each focus from the centre, and
• the length of the transverse axis is $$2a$$, so $$a$$ is half that length.

Step 1 – Identify $$c$$

The given foci are $$(\pm 5,0)$$. Therefore

$$c = 5 \;\;\;\Rightarrow\;\;\; c^{2} = 25$$

Step 2 – Find $$a$$ from the transverse axis length

Length of the transverse axis = 8, hence

$$2a = 8 \;\;\;\Rightarrow\;\;\; a = 4 \;\;\;\Rightarrow\;\;\; a^{2} = 16$$

Step 3 – Compute $$b^{2}$$ using the relation $$c^{2}=a^{2}+b^{2}$$

$$b^{2} = c^{2} - a^{2} = 25 - 16 = 9$$

Step 4 – Write the equation of the hyperbola

Substituting $$a^{2}=16$$ and $$b^{2}=9$$ in the standard form, we get

$$\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$$

Step 1 — Identify the orientation and the unknowns
Because the two foci are on the y-axis at $$(0,\,\pm 13)$$, the centre is the origin and the transverse axis is vertical.
For a hyperbola whose transverse axis is vertical we use the standard form

$$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$$

Here a, b, c satisfy the fundamental relation

$$c^2=a^2+b^2$$

and

• co-ordinates of the foci: $$(0,\,\pm c)$$,
• length of the conjugate axis: $$2b$$.

Step 2 — Insert the given data

• Given foci $$(0,\,\pm 13)\;\Rightarrow\;c=13$$.
• Conjugate axis length $$24\;\Rightarrow\;2b=24\;\Rightarrow\;b=12$$.

Step 3 — Compute $$a^2$$ from the relation $$c^2=a^2+b^2$$

$$a^2=c^2-b^2=13^2-12^2=169-144=25$$

Step 4 — Write the required equation

Inserting $$a^2=25$$ and $$b^2=144$$ into the standard form gives

$$\frac{y^2}{25}-\frac{x^2}{144}=1$$

Thus the hyperbola that meets all the given conditions is

$$\boxed{\displaystyle\frac{y^2}{25}-\frac{x^2}{144}=1}$$

Step 1 : Choose the standard form
Because the foci are on the x-axis (the points are $$(\pm 3\sqrt{5},0)$$), the transverse axis is horizontal. The hyperbola is therefore taken in its centre–at–origin, transverse-axis-along-x form

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \;\;(a>0,\,b>0).$$

For this form we have the summary relations

• Foci : $$(\pm c,0)$$ with $$c^{2}=a^{2}+b^{2}$$.
• Length of a latus-rectum : $$\displaystyle \ell =\frac{2b^{2}}{a}$$.

Step 2 : Insert the given focus information
The foci are $$(\pm 3\sqrt{5},0)$$, hence

$$c = 3\sqrt{5}\quad\Longrightarrow\quad c^{2}=45.$$

Step 3 : Insert the given latus-rectum length
We are told that $$\ell =8$$, so

\[\frac{2b^{2}}{a}=8\quad\Longrightarrow\quad b^{2}=4a.\tag{1}\]

Step 4 : Use the focus–parameter relation

\[c^{2}=a^{2}+b^{2}\quad\Longrightarrow\quad 45=a^{2}+b^{2}.\tag{2}\]

Step 5 : Solve simultaneously for a and b
Substitute the expression for $$b^{2}$$ from (1) into (2):

$$45=a^{2}+4a\;\;\Longrightarrow\;\;a^{2}+4a-45=0.$$

This quadratic gives

$$a=\frac{-4\pm\sqrt{(-4)^{2}-4(1)(-45)}}{2} =\frac{-4\pm\sqrt{16+180}}{2} =\frac{-4\pm14}{2}.$$

The positive root is required (since $$a>0$$):

$$a=\frac{-4+14}{2}=\frac{10}{2}=5.$$

Now find $$b^{2}$$ from (1):

$$b^{2}=4a=4\times5=20.$$

Step 6 : Write the equation

$$\boxed{\displaystyle \frac{x^{2}}{5}-\frac{y^{2}}{20}=1}. $$

This is the required hyperbola whose foci are $$(\pm3\sqrt{5},0)$$ and whose latus-rectum has length 8.

Step 1  Identify the suitable standard form
Because the foci are on the x-axis, the transverse axis is the x-axis and the required hyperbola is of the form
\[\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1\tag{1}\]
For this curve

• Centre : $$O(0,0)$$
• Foci : $$F_1(-c,0),\;F_2(c,0)$$ where $$c>0$$
• Latus-rectum length : $$\displaystyle 2\frac{b^{2}}{a}$$
• Relation : $$c^{2}=a^{2}+b^{2}$$

Step 2  Use the given foci
Given foci $$(\pm4,0)$$ give
$$c = 4$$

Step 3  Use the given length of the latus rectum
Length $=12$:
\[2\frac{b^{2}}{a}=12 \;\Longrightarrow\; b^{2}=6a\tag{2}\]

Step 4  Relate a, b, c
Insert $$c=4$$ and (2) in $$c^{2}=a^{2}+b^{2}$$:
$$16=a^{2}+b^{2}=a^{2}+6a$$
$$\Rightarrow\; a^{2}+6a-16=0$$

Step 5  Solve for a
$$a=\frac{-6\pm\sqrt{36+64}}{2}=\frac{-6\pm10}{2}$$
The positive root is required, hence $$a=2$$

Step 6  Find b²
From (2): $$b^{2}=6a=6\times2=12$$

Step 7  Write the equation
Substitute $$a^{2}=4,\;b^{2}=12$$ in (1):
$$\boxed{\dfrac{x^{2}}{4}-\dfrac{y^{2}}{12}=1}$$

This hyperbola has foci $$(\pm4,0)$$ and a latus-rectum of length $$12$$, satisfying the given conditions.

Let the required hyperbola be centred at the origin.
The given vertices are $$(7,0)$$ and $$(-7,0)$$; hence

$$a=7,\text{ so }a^{2}=49.$$

The eccentricity is $$e=\dfrac{4}{3}$$. For a hyperbola, $$e=\dfrac{c}{a}$$, where $$c$$ is the distance from the centre to a focus. Therefore

$$c=e\,a=\dfrac{4}{3}\times 7=\dfrac{28}{3},\qquad c^{2}=\left(\dfrac{28}{3}\right)^{2}=\dfrac{784}{9}.$$

For the rectangular hyperbola oriented along the $$x$$-axis, the relation between $$a$$, $$b$$ and $$c$$ is

$$c^{2}=a^{2}+b^{2}\;\Rightarrow\;b^{2}=c^{2}-a^{2}.$$

Substituting the known values:

$$b^{2}=\dfrac{784}{9}-49=\dfrac{784}{9}-\dfrac{441}{9}=\dfrac{343}{9}.$$

Hence $$b^{2}=\dfrac{343}{9}$$.

The standard equation of a hyperbola with its transverse axis along the $$x$$-axis is

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1.$$

Substituting $$a^{2}=49$$ and $$b^{2}=\dfrac{343}{9}$$ gives

$$\frac{x^{2}}{49}-\frac{y^{2}}{\tfrac{343}{9}}=1.$$

To remove the complex fraction, multiply numerator and denominator of the second term by $$9$$:

$$\frac{x^{2}}{49}-\frac{9y^{2}}{343}=1.$$

Either form represents the required hyperbola.

Let the centre of the hyperbola be the origin O(0,0). Since the two foci are given as $$(0,\,\pm\sqrt{10})$$, the transverse axis is the y-axis. Hence the required hyperbola is of the standard form

$$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1 \quad (a>0,\,b>0).$$

For a hyperbola with this orientation, $$c^{2}=a^{2}+b^{2}$$, where $$c$$ is the focal distance. Here $$c^{2}=10$$ (because the foci are at $$\pm\sqrt{10}$$ on the y-axis). Therefore

\[a^{2}+b^{2}=10. \tag{1}\]

The hyperbola passes through the point $$(2,3)$$, so its coordinates satisfy the equation:

\[\frac{(3)^{2}}{a^{2}}-\frac{(2)^{2}}{b^{2}}=1 \;\Longrightarrow\; \frac{9}{a^{2}}-\frac{4}{b^{2}}=1.\tag{2}\]

Put $$a^{2}=A$$ and $$b^{2}=B$$ to simplify (1) and (2):

From (2):  \[\dfrac{9}{A}-1=\dfrac{4}{B}.\tag{3}\]

Using (1), $$B=10-A$$. Substitute this in (3):

\[\frac{9}{A}-1=\frac{4}{10-A} \;\Longrightarrow\; \frac{9-A}{A}=\frac{4}{10-A}.\tag{4}\]

Cross-multiply:

$$(9-A)(10-A)=4A.$$

Expand and collect terms:

$$90-19A+A^{2}=4A \;\Longrightarrow\; A^{2}-23A+90=0.$$

Solve the quadratic:

$$A=\frac{23\pm\sqrt{23^{2}-4\cdot90}}{2}=\frac{23\pm13}{2}.$$

This gives $$A=18$$ or $$A=5$$. But if $$A=18$$, then from (1) $$B=10-18=-8<0$$, impossible because $$b^{2}$$ cannot be negative. Hence

$$a^{2}=A=5, \qquad b^{2}=B=10-5=5.$$

Substitute these values into the standard form:

$$\frac{y^{2}}{5}-\frac{x^{2}}{5}=1.$$

Multiplying through by 5 gives the simplest form

$$y^{2}-x^{2}=5.$$

Step 1 : Choose a coordinate system

Place the vertex of the mirror at the origin $$O(0,0)$$ and take the axis of the mirror along the positive $$x$$-axis; the parabola therefore opens to the right.

Step 2 : Equation of the parabola

For a right-opening parabola with vertex at the origin and focus at $$(a,0)$$ the standard form is $$y^{2}=4ax$$.

The focus is $$5\,\mathrm{cm}$$ from the vertex, so $$a = 5$$.

Hence $$y^{2}=4(5)x \;\Rightarrow\; y^{2}=20x.$$

Step 3 : Coordinate of the rim

The mirror is $$45\,\mathrm{cm}$$ deep, i.e. the rim lies on the line $$x = 45$$.

Step 4 : Points $$A$$ and $$B$$ on the rim

Putting $$x = 45$$ in $$y^{2}=20x$$ gives

$$y^{2}=20\times 45 = 900 \;\Rightarrow\; y = \pm 30.$$

Thus $$A(45,30)$$ and $$B(45,-30).$$

Step 5 : Distance $$AB$$

$$AB = |30-(-30)| = 60\,\text{cm}.$$

Therefore, the required distance is $${60\,\text{cm}}.$$

Step 1 : Choose a convenient coordinate system

• The unloaded beam is taken as the $$x$$–axis (so its equation is $$y = 0$$).
• The two supports are $$12\,\text{m}=1200\,\text{cm}$$ apart; place them at $$x=-600\,\text{cm}$$ and $$x=+600\,\text{cm}$$.
• The load is at the centre, so the lowest point of the deflected beam is exactly midway between the supports. Hence the vertex of the parabola is at the origin of $$x$$ but $$3\,\text{cm}$$ below the $$x$$–axis. Thus the vertex is the point $$(0,-3)$$.

Step 2 : Write the equation of the parabola

With vertex at $$(0,-3)$$ and axis along the $$y$$–direction, the equation is in vertex form

$$y+3 = a x^2 \;\;\Longrightarrow\;\; y = a x^2 - 3$$

Step 3 : Determine the value of $$a$$ using a support point

The parabola passes through the right-hand support $$(600,0)$$:

$$0 = a(600)^2 - 3 \;\Longrightarrow\;\; a = \dfrac{3}{600^2}=\dfrac{3}{360000}=\dfrac{1}{120000}$$

Hence the required equation is

$$y = \dfrac{1}{120000}\,x^2 - 3$$

Step 4 : Locate the points where the deflection is $$1\,\text{cm}$$

A downward deflection of $$1\,\text{cm}$$ means $$y = -1$$.

Substitute $$y = -1$$:

$$-1 = \dfrac{1}{120000}\,x^2 - 3$$
$$\Rightarrow\; -1 + 3 = \dfrac{x^2}{120000}$$
$$\Rightarrow\; 2 = \dfrac{x^2}{120000}$$
$$\Rightarrow\; x^2 = 2 \times 120000 = 240000$$
$$\Rightarrow\; x = \pm\sqrt{240000}\;\text{cm}$$

Simplify:

$$\sqrt{240000}=\sqrt{24\times10000}=100\sqrt{24}=100\times2\sqrt{6}=200\sqrt{6}\;\text{cm}$$

Step 5 : Convert to metres

$$200\sqrt{6}\;\text{cm}= \dfrac{200\sqrt{6}}{100}\;\text{m}=2\sqrt{6}\;\text{m}\approx4.90\;\text{m}$$

Thus the deflection is $$1\,\text{cm}$$ at a distance of $$2\sqrt{6}\,\text{m}\;(\approx4.9\,\text{m})$$ on either side of the centre.

Let the coordinates of the end–points of the rod be

• $$A(a,0)$$   (on the $$x$$–axis)
• $$B(0,b)$$   (on the $$y$$–axis), where $$a>0,\;b>0$$

Because the length of the rod is $$AB = 15\,\text{cm}$$,

\[\sqrt{a^{2}+b^{2}} = 15 \;\;\Longrightarrow\;\; a^{2}+b^{2}=225 \tag{1}\]

The point $$P(x,y)$$ is on the rod such that $$AP = 6\,\text{cm}$$.
Since $$AB = 15\,\text{cm}$$,

$$\dfrac{AP}{AB}=\dfrac{6}{15}=\dfrac{2}{5}$$.

Hence $$P$$ divides $$AB$$ internally in the ratio $$AP:PB = 2:3$$ (because $$AB=AP+PB\Rightarrow PB=15-6=9\text{ cm}$$).

By the section formula,

\[\begin{aligned} x &= \dfrac{3\,a + 2\,0}{3+2} = \dfrac{3a}{5},\\[4pt] y &= \dfrac{3\,0 + 2\,b}{3+2} = \dfrac{2b}{5}.\tag{2} \end{aligned}\]

Express $$a$$ and $$b$$ from (2):

\[a = \dfrac{5x}{3},\qquad b = \dfrac{5y}{2}.\tag{3}\]

Substitute (3) in (1):

$$\left(\dfrac{5x}{3}\right)^{2}+\left(\dfrac{5y}{2}\right)^{2}=225$$ $$\Longrightarrow\; \dfrac{25x^{2}}{9}+\dfrac{25y^{2}}{4}=225$$ $$\Longrightarrow\; \dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=9$$ \[\Longrightarrow\; \dfrac{x^{2}}{81}+\dfrac{y^{2}}{36}=1.\tag{4}\]

Equation (4) is of the standard form $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$ with $$a^{2}=81,\;b^{2}=36$$, i.e.

$$a = 9\text{ cm},\; b = 6\text{ cm}.$$

Since both denominators are positive and unequal, (4) represents an ellipse centred at the origin, having semi-major axis $$9\,\text{cm}$$ along the $$x$$-axis and semi-minor axis $$6\,\text{cm}$$ along the $$y$$-axis.

Therefore, the locus of the point $$P$$ is the ellipse

$$\dfrac{x^{2}}{81}+\dfrac{y^{2}}{36}=1.$$

Step 1 : Choose a convenient coordinate system
Place the reflector so that its rim is horizontal and its deepest point (the bottom of the dish) is the origin O (0,0).
The axis of the dish is the y-axis, the dish opens upward and its cross-section is a parabola.

Step 2 : Write the general equation
For a parabola with vertex at the origin and axis along the positive y-direction, the simplest form is
$$y = a x^{2}$$ where $$a > 0$$.

Step 3 : Use the given diameter and depth
The diameter of the rim is $$20\,\mathrm{cm}$$, so the rim points are $$10\,\mathrm{cm}$$ to the left and right of the axis.
The depth is $$5\,\mathrm{cm}$$, so these rim points lie $$5\,\mathrm{cm}$$ above the vertex.

Thus the point $$P(10,5)$$ lies on the parabola. Substitute it in $$y = a x^{2}$$:

$$5 = a(10)^{2} \;\;\Longrightarrow\;\; 5 = 100 a \;\;\Longrightarrow\;\; a = \frac{5}{100} = \frac{1}{20}.$$

Step 4 : Relate the coefficient a to the focal length
Rewrite $$y = ax^{2}$$ as $$x^{2} = \frac{1}{a} \, y$$.
The standard focus form is $$x^{2} = 4 p \, y$$, where $$p$$ is the focal length (the distance from the vertex to the focus).

Matching coefficients: $$4p = \frac{1}{a} \;\;\Longrightarrow\;\; p = \frac{1}{4a}.$$

Substitute $$a = \frac{1}{20}$$:

$$p = \frac{1}{4 \times \frac{1}{20}} = \frac{1}{\frac{1}{5}} = 5.$$

Step 5 : Interpret the result
The focus lies on the axis of symmetry, $$5\,\mathrm{cm}$$ above (inside) the vertex O.
Hence a light source placed $$5\,\mathrm{cm}$$ in front of the dish’s bottom will produce parallel reflected rays.

Step 1 : Choose convenient axes

• Place the vertex of the arch at the origin $$O(0,0)$$.
• Take the positive $$y$$–axis vertically upward; hence the arch, which opens downward, will have negative $$y$$–values.
• The $$x$$–axis is the horizontal line through the vertex; the arch is symmetric about the $$y$$–axis.

With this choice the equation of the arch is of the form $$y=-a x^{2}$$, where $$a>0$$.

Step 2 : Use the information at the base of the arch

The arch is $$10\,\mathrm{m}$$ high, so the base (where it meets the ground) is $$10\,\mathrm{m}$$ below the vertex. At that level the total width is $$5\,\mathrm{m}$$. Hence the two end-points of the arch are

$$A\bigl(-2.5,-10\bigr),\;B\bigl(2.5,-10\bigr).$$

Substituting one of these points in $$y=-a x^{2}$$ gives

$$\begin{aligned}-10 & = -a(2.5)^{2} \\[2pt]a & = \frac{10}{(2.5)^{2}}=\frac{10}{6.25}=1.6.\end{aligned}$$

Step 3 : Write the explicit equation of the arch

Therefore the parabola is

$$y = -1.6x^{2}.$$

Step 4 : Find the width at a point $$2\,\mathrm{m}$$ below the vertex

“$$2\,\mathrm{m}$$ from the vertex” along the axis means $$y=-2$$. Substitute $$y=-2$$ into the equation:

$$\begin{aligned}-2 & = -1.6x^{2} \\[2pt]x^{2} & = \frac{2}{1.6}=\frac{5}{4}=1.25 \\[2pt]x & = \pm\sqrt{1.25}=\pm\frac{\sqrt{5}}{2}\;\text{metres}.\end{aligned}$$

Step 5 : Compute the required width

The arch is symmetric, so the width at that level is

$$\text{width}=2\left(\frac{\sqrt{5}}{2}\right)=\sqrt{5}\,\text{metres}\approx2.24\,\text{m}.$$

Conclusion

The arch is approximately $$2.24\,\text{m}$$ wide at a point $$2\,\text{m}$$ below the vertex.

Step 1 – Choose a coordinate system
Take the straight roadway as the X–axis and its mid-point as the origin O(0, 0).
The Y–axis is vertical and positive upwards.
Because the cables are above the roadway, the length of every supporting wire equals the Y-coordinate of the cable at the same X-coordinate.

Step 2 – Write the general equation of the cable
The cable is symmetric about the Y-axis, so its equation can be written as
$$y = ax^{2} + c$$ with a, c constants to be found.

Step 3 – Translate the data into coordinate pairs

• At the centre of the span (x = 0) the wire is the shortest, 6 m long:
  $$y(0)=6\Rightarrow c = 6.$$
• The span is 100 m long, so the ends of the roadway are 50 m either side of the origin (x = ±50). At each end the wire is the longest, 30 m:
  $$y(\pm 50)=30\Rightarrow a(50)^{2}+6 = 30.$$ Hence $$a = \dfrac{30-6}{50^{2}} = \dfrac{24}{2500}=\dfrac{12}{1250}.$$

Step 4 – Find the length of the wire 18 m from the middle
The required point is x = 18. Substitute:

$$\begin{aligned} y(18) &= \dfrac{12}{1250}(18)^{2}+6\\ &= \dfrac{12}{1250}\times 324 + 6\\ &= \dfrac{3888}{1250}+6\\ &= \dfrac{3888+7500}{1250}\\ &= \dfrac{11388}{1250}=\dfrac{5694}{625}\;\text{m}\\ &\approx 9.11\;\text{m}. \end{aligned}$$

Step 5 – Interpretation
Therefore the supporting wire fixed 18 m from the centre of the bridge must be about 9.11 metres long.

Step 1 : Choose a convenient coordinate system
Let the arch rest on the horizontal ground.
Take the mid-point of the base as origin O.
• The x-axis lies along the ground with the positive direction to the right.
• The y-axis is vertical, measured upwards from the ground.

Step 2 : Write the equation of the ellipse
The arch is a semi-ellipse, so its full ellipse is symmetric about both axes.
Width = $$8\,\text{m} \;\Rightarrow\; 2a = 8 \;\Rightarrow\; a = 4$$.
Height at the centre = $$2\,\text{m} \;\Rightarrow\; b = 2$$.
Hence the equation of the complete ellipse is
$$\frac{x^{2}}{4^{2}} + \frac{y^{2}}{2^{2}} = 1 \;\;\Longrightarrow\;\; \frac{x^{2}}{16} + \frac{y^{2}}{4} = 1.$$

Step 3 : Locate the required point on the base
"$$1.5\,\text{m}$$ from one end" means a horizontal distance of $$1.5\,\text{m}$$ from an end point.
The end points of the base are at $$x = \pm 4$$.
Taking the left end ( $$x = -4$$ ) and moving $$1.5\,\text{m}$$ to the right gives
$$x = -4 + 1.5 = -2.5.$$
Because the ellipse is symmetric, we may use $$|x| = 2.5$$.
Hence take $$x = 2.5\,(\text{m}).$$

Step 4 : Find the corresponding height
Substitute $$x = 2.5$$ in the ellipse equation:
$$\frac{(2.5)^{2}}{16} + \frac{y^{2}}{4} = 1.$$
Compute:
$$\frac{6.25}{16} + \frac{y^{2}}{4} = 1\;\;\Longrightarrow\;\; \frac{y^{2}}{4} = 1 - 0.390625 = 0.609375.$$
Multiply by $$4$$:
$$y^{2} = 4 \times 0.609375 = 2.4375.$$
Take the positive square root (upper half of the ellipse):
$$y = \sqrt{2.4375} \approx 1.5612495\,\text{m}.$$

Step 5 : State the result
Therefore the height of the arch $$1.5\,\text{m}$$ from either end is approximately $$1.56\,\text{m}$$ (to two decimal places).

Step 1 : Choose coordinates for the ends of the rod
Let the end on the x-axis be $$A(a,0)$$ and the end on the y-axis be $$B(0,b)$$. Since the rod is $$12\,\text{cm}$$ long, the distance formula gives

$$AB^{2}=a^{2}+b^{2}=12^{2}=144\;.$$

Step 2 : Locate the required point P on the rod
Point $$P$$ is $$3\,\text{cm}$$ from $$A$$, so $$AP:PB=3:9=1:3$$ (because the whole rod is $$12\,\text{cm}$$).
Using the internal-division (section) formula, if $$P(x,y)$$ divides $$A(a,0)$$ and $$B(0,b)$$ in the ratio $$1:3$$, then

$$x=\frac{(3)\,a+(1)\,0}{1+3}=\frac{3a}{4},\qquad y=\frac{(3)\,0+(1)\,b}{1+3}=\frac{b}{4}\;.$$

Step 3 : Express $$a$$ and $$b$$ through $$x$$ and $$y$$
From the two relations just obtained,

$$a=\frac{4x}{3},\qquad b=4y\;.$$

Step 4 : Eliminate $$a$$ and $$b$$ using the length condition
Substitute $$a=\dfrac{4x}{3}$$ and $$b=4y$$ into $$a^{2}+b^{2}=144$$:

$$\left(\frac{4x}{3}\right)^{2}+(4y)^{2}=144\;.$$

Simplify:

$$\frac{16x^{2}}{9}+16y^{2}=144.\;$$

Multiply by $$9$$ to clear the denominator:

$$16x^{2}+144y^{2}=1296.\;$$

Divide by $$16$$:

$$x^{2}+9y^{2}=81.\;$$

Step 5 : Write the equation in standard form
Dividing both sides by $$81$$ gives the standard form of an ellipse:

$$\frac{x^{2}}{81}+\frac{y^{2}}{9}=1.\;$$

Hence, the locus of the point $$P$$ is the ellipse
$$\boxed{\dfrac{x^{2}}{81}+\dfrac{y^{2}}{9}=1}. $$

Step 1 : Write the parabola in standard form
Given $$x^2 = 12y$$. Comparing with the standard form $$x^2 = 4ay$$ we obtain $$4a = 12 \;\Rightarrow\; a = 3$$.

Step 2 : Vertex, focus and latus rectum
The vertex is $$V(0,0)$$. Since $$a = 3$$, the focus is $$S(0,3)$$ and the length of the latus-rectum is $$4a = 12$$. The latus-rectum is the line through the focus perpendicular to the axis (the axis is the $$y$$–axis), therefore its equation is $$y = 3$$.

Step 3 : Endpoints of the latus-rectum
To find the endpoints we substitute $$y = 3$$ in the equation of the parabola:
$$x^2 = 12(3) = 36 \;\Rightarrow\; x = \pm 6$$.
Hence the endpoints are $$L(6,3)$$ and $$L'(-6,3)$$.

Step 4 : Triangle formed with the vertex
The required triangle has vertices
$$V(0,0),\; L(6,3),\; L'(-6,3).$$

Step 5 : Area of the triangle

• Method I : Base–height
  The segment $$LL'$$ is horizontal, so its length is the base: $$|6-(-6)| = 12$$.
  The perpendicular distance of $$V$$ from the line $$y=3$$ is simply $$3$$.
  Area $$= \dfrac12 \times \text{base} \times \text{height} = \dfrac12 \times 12 \times 3 = 18.$$
• Method II : Determinant formula (check)
  $$\text{Area} = \dfrac12 \left| \begin{vmatrix}0 & 0 & 1\\ 6 & 3 & 1\\ -6 & 3 & 1\end{vmatrix} \right| = \dfrac12\,18 = 18.$$

Therefore, the area of the required triangle is $$18\;\text{square units}.$$

Step 1 : Set up a convenient coordinate system
Choose the origin $$O(0,0)$$ to be the midpoint of the two flag-posts, and let the line joining the posts be the $$x$$-axis.
Hence the coordinates of the posts (foci) are $$F_1(-c,0)$$ and $$F_2(c,0)$$.

Step 2 : Insert the numerical data
The distance between the two posts is given to be $$8\;\text{m}$$.
Therefore
$$2c = 8 \;\Rightarrow\; c = 4 \;\text{m}$$

Step 3 : Translate the condition on distances
A point $$P(x,y)$$ represents the runner’s instantaneous position. We are told that the sum of its distances from the two posts is constantly $$10\;\text{m}$$, that is
$$PF_1 + PF_2 = 10$$
By definition, the locus of a point whose sum of distances from two fixed points is constant is an ellipse. For such an ellipse, the constant sum equals the length of the major axis, denoted $$2a$$. Hence
$$2a = 10 \;\Rightarrow\; a = 5 \;\text{m}$$

Step 4 : Find the semi-minor axis
For every ellipse with centre at the origin and foci on the $$x$$-axis we have the relation
$$c^2 = a^2 - b^2$$
Substituting $$a = 5$$ and $$c = 4$$ gives
$$b^2 = a^2 - c^2 = 25 - 16 = 9 \;\Longrightarrow\; b = 3\;\text{m}$$

Step 5 : Write the standard equation
Because the major axis lies on the $$x$$-axis, the standard form is
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Putting $$a^2 = 25$$ and $$b^2 = 9$$ we obtain
$$\frac{x^2}{25} + \frac{y^2}{9} = 1$$

Result
The runner traces the ellipse
$$\displaystyle \frac{x^2}{25} + \frac{y^2}{9} = 1$$ with its centre at the midpoint of the two flag-posts, major axis $$10\;\text{m}$$ long, and foci $$8\;\text{m}$$ apart.

Let the given parabola be $$y^{2}=4ax$$ with vertex at the origin $$O(0,0)$$.
Take this point as one vertex of the required equilateral triangle and call it $$A(0,0)$$.

Any other point on the parabola can be written in parametric form as
$$P(at^{2},\,2at)\;\; (t\in\mathbb R).$$

Because the triangle has to be equilateral and the axis of the parabola is the x-axis, it is natural (and will soon be confirmed) to choose the remaining two vertices symmetrically with respect to that axis:

• $$B(at^{2},\,2at)$$
• $$C(at^{2},\,-2at)$$

Thus
$$A(0,0),\; B(at^{2},2at),\; C(at^{2},-2at).$$

1. Compute the three side lengths.

AB (and AC, by symmetry):
$$\begin{aligned} AB&=\sqrt{(at^{2}-0)^{2}+(2at-0)^{2}}\\ &=\sqrt{a^{2}t^{4}+4a^{2}t^{2}}\\ &=a|t|\sqrt{t^{2}+4}. \end{aligned}$$

BC (vertical segment):
$$BC=|2at-(-2at)|=4at\;\;(t>0\;\text{to avoid repetition of the vertex}).$$

2. Impose the equilateral‐triangle condition.
For an equilateral triangle we need $$AB=BC$$. Taking $$t>0$$ so that $$|t|=t$$:

Square both sides:

3. Evaluate the common side length.

Using $$t=2\sqrt{3}$$ in any side (say $$BC$$):

The same value is obtained for $$AB$$ and $$AC$$, so the triangle is indeed equilateral.

Hence the length of each side of the equilateral triangle is $$8\sqrt{3}\,a$$.