Standard Equation of a Parabola

Rarely Tested

$$y^2=4ax$$

Terminology

- $$a$$ → Distance of the focus from the vertex.

- Vertex → $$(0,0)$$.

- Focus → $$(a,0)$$.

- Axis → $$x$$-axis.

- Directrix → $$x=-a$$.

Usage

- Used for a parabola opening towards the positive $$x$$-axis.

Question 1

Let $$T$$ be the tangent to the parabola $$y^2=16x$$ at the point $$(64,32)$$. Let $$L$$ be the tangent to the same parabola at another point $$(x_1,\,y_1)$$ on the parabola. If $$L$$ and $$T$$ are perpendicular to each other, then the distance between the point $$(x_1,\,y_1)$$ and the focus of the parabola, is

Question 2

A point $$P$$ lies on the parabola $$y^2 = 8x$$ such that its ordinate (y-coordinate) is exactly twice its abscissa (x-coordinate). If a focal chord of the parabola passes through this point $$P$$, find the total length of this focal chord.

Question 3

A normal chord of the parabola $$y^2=8x$$ at a point $$P$$ in the first quadrant subtends a right angle at

the focus of the parabola. Find the $$x$$ -coordinate of $$P$$.

Solution

The given parabola is $$y^{2}=8x$$.
Rewrite it in the standard form $$y^{2}=4ax$$ to identify $$a$$:

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

For $$y^{2}=4ax$$ a convenient parametric point is
$$P\,(x,y)=\bigl(a\,t^{2},\,2a\,t\bigr)=\bigl(2t^{2},\,4t\bigr)$$ with $$t\gt 0$$ (first quadrant).

Normal at $$P$$
Slope of the tangent at parameter $$t$$ is $$\dfrac{1}{t}$$, so the slope of the normal is $$-t$$.
Equation of the normal through $$P(2t^{2},4t)$$ is

$$y-4t=-t\bigl(x-2t^{2}\bigr) \;\;\Longrightarrow\;\; y=-tx+4t+2t^{3} \;-(1)$$

The second intersection of this normal with the parabola (besides $$P$$) is $$Q$$.
Substitute $$y$$ from $$(1)$$ into $$y^{2}=8x$$ to locate $$Q$$:

$$( -t x + 4t + 2t^{3})^{2}=8x$$

This is a quadratic in $$x$$ whose two roots are $$x_{P}=2t^{2}$$ and $$x_{Q}$$.
Using the sum of roots formula $$\dfrac{-b}{a}$$:

$$x_{P}+x_{Q}=8+4t^{2}+\dfrac{8}{t^{2}} \;\;\Longrightarrow\;\; x_{Q}=8+2t^{2}+\dfrac{8}{t^{2}} \;-(2)$$

Coordinates of $$Q$$
Insert $$x_{Q}$$ from $$(2)$$ into the normal $$(1)$$:

$$y_{Q}= -t\!\left(8+2t^{2}+\dfrac{8}{t^{2}}\right)+4t+2t^{3} =-4t-\dfrac{8}{t} \;-(3)$$

Right-angle condition at the focus
Focus of the parabola is $$S(2,0)$$.
Vectors $$\overrightarrow{SP}$$ and $$\overrightarrow{SQ}$$ must be perpendicular, i.e. their dot product is zero:

$$\overrightarrow{SP}=(2t^{2}-2,\,4t)$$
$$\overrightarrow{SQ}=\bigl(x_{Q}-2,\,y_{Q}\bigr) =\left(6+2t^{2}+\dfrac{8}{t^{2}},\,-4t-\dfrac{8}{t}\right)$$

Dot product:

$$(2t^{2}-2)\!\left(6+2t^{2}+\dfrac{8}{t^{2}}\right)+4t\!\left(-4t-\dfrac{8}{t}\right)=0$$

Simplifying gives

$$4t^{6}-8t^{4}-28t^{2}-16=0$$

Divide by 4 and put $$u=t^{2}$$ (since $$t\gt0$$):

$$u^{3}-2u^{2}-7u-4=0$$

This factors as $$\bigl(u-4\bigr)\bigl(u+1\bigr)^{2}=0$$, so the positive solution is

$$u=4\;\Longrightarrow\; t^{2}=4\;\Longrightarrow\; t=2$$

Required $$x$$-coordinate of $$P$$

$$x_{P}=2t^{2}=2(4)=8$$

Hence, the point $$P$$ has $$x$$-coordinate $$8$$.

Option A which is: $$8$$

Question 4

A parabola with vertex at the origin and axis along the positive x-axis passes through the point $$(4, 4\sqrt{2})$$. A circle with center at the origin intersects this parabola at two points $$P$$ and $$Q$$ such that the chord $$PQ$$ subtends a right angle at the vertex. Find the radius of the circle.

Solution

The parabola has its vertex at the origin and axis along the positive $$x$$-axis, so its standard form is $$y^{2}=4ax$$.

Because the point $$(4,\,4\sqrt2)$$ lies on the parabola, substitute to find $$a$$:
$$\bigl(4\sqrt2\bigr)^2=4a(4)\;\;\Longrightarrow\;\;32=16a\;\;\Longrightarrow\;\;a=2.$$

Hence the parabola is
$$y^{2}=8x\quad -(1)$$

Let the required circle be $$x^{2}+y^{2}=R^{2} \; -(2).$$ (The centre is also at the origin; we denote the radius by $$R$$.)

Using the parametric coordinates of the parabola $$x=2t^{2},\qquad y=4t$$ (where $$t$$ is any real parameter) we impose condition $$-(2)$$:

$$\bigl(2t^{2}\bigr)^{2}+\bigl(4t\bigr)^{2}=R^{2} \;\;\Longrightarrow\;\;4t^{4}+16t^{2}=R^{2} \;\;\Longrightarrow\;\;R^{2}=4t^{2}\bigl(t^{2}+4\bigr). \quad -(3)$$

If the circle meets the parabola at points $$P(t_1)$$ and $$Q(t_2)$$, then both parameters $$t_1,t_2$$ satisfy $$-(3)$$ with the same $$R$$.

The chord $$PQ$$ subtends a right angle at the vertex, i.e. the origin. So the vectors $$\overrightarrow{OP}$$ and $$\overrightarrow{OQ}$$ are perpendicular:

$$\overrightarrow{OP}\cdot\overrightarrow{OQ}=0.$$

Now $$\overrightarrow{OP}=\bigl(2t_1^{2},\,4t_1\bigr),\qquad \overrightarrow{OQ}=\bigl(2t_2^{2},\,4t_2\bigr).$$

Taking their dot product, $$\bigl(2t_1^{2}\bigr)\bigl(2t_2^{2}\bigr)+\bigl(4t_1\bigr)\bigl(4t_2\bigr)=4t_1^{2}t_2^{2}+16t_1t_2 =4t_1t_2\bigl(t_1t_2+4\bigr)=0.$$

Since neither point coincides with the origin, $$t_1t_2\neq0$$, hence $$t_1t_2+4=0\;\;\Longrightarrow\;\;t_1t_2=-4. \quad -(4)$$

Using $$-(3)$$ for both parameters: $$R^{2}=4t_1^{2}(t_1^{2}+4)=4t_2^{2}(t_2^{2}+4).$$ Replace $$t_2$$ by $$-4/t_1$$ from $$-(4)$$ and equate the two expressions:

$$4t_1^{2}(t_1^{2}+4)=4\!\left(\frac{16}{t_1^{2}}\right)\!\left(\frac{16}{t_1^{2}}+4\right).$$

Simplifying (let $$u=t_1^{2}\;(\,u\gt0\,)$$): $$u^{4}+4u^{3}-64u-256=0.$$ This polynomial factors as $$(u-4)(u+4)\bigl(u^{2}+4u+16\bigr)=0,$$ yielding the only positive root $$u=4.$$

Therefore $$t_1^{2}=4\;\;\Longrightarrow\;\;t_1=\pm2,$$ and by $$-(4)$$ $$t_2=-\frac{4}{t_1}=\mp2.$$ Thus the two parameters are $$t_1=2,\;t_2=-2$$ (or vice-versa).

Finally, substitute $$t_1=2$$ (or $$t_2=-2$$) into $$-(3)$$ to get the radius:

$$R^{2}=4(2)^{2}\bigl((2)^{2}+4\bigr)=4\cdot4\cdot8=128 \;\;\Longrightarrow\;\;R=\sqrt{128}=8\sqrt2.$$

Hence the radius of the circle is $$8\sqrt2$$.

Option C which is: $$8\sqrt2$$

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Standard Equation of a Parabola

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