Focus of $$y^2=4ax$$

Step 1: Find the Focus of the Parabola

The standard equation of a rightward-opening parabola is $$y^2 = 4ax$$.

Comparing this to our given equation:

$$y^2 = 12x \implies 4a = 12 \implies a = 3$$

The focus of the parabola is located at $$(a, 0)$$, which is $$(3, 0)$$.

Step 2: Set up the Equation of the Focal Chord

Any line passing through the focus $$(3, 0)$$ with a slope $$m$$ can be written using the point-slope form:

$$y - 0 = m(x - 3)$$  $$mx - y - 3m = 0$$

Step 3: Apply the Condition of Tangency to the Circle

The given equation of the circle is:

$$(x - 5)^2 + y^2 = 2$$

From this standard form, we can identify:

• Center ($$C$$): $$(5, 0)$$
• Radius ($$R$$): $$\sqrt{2}$$

For the line to be a tangent to the circle, the perpendicular distance from the center $$(5, 0)$$ to the line must equal the radius $$R$$:

$$\frac{|m(5) - (0) - 3m|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$$

Simplify the expression inside the absolute value:

$$\frac{|2m|}{\sqrt{m^2 + 1}} = \sqrt{2}$$

Step 4: Solve for the Slopes ($$m$$)

Square both sides of the equation to eliminate the modulus and radical signs:

$$\frac{4m^2}{m^2 + 1} = 2$$

Cross-multiply and simplify:

$$4m^2 = 2(m^2 + 1)$$  $$4m^2 = 2m^2 + 2$$  $$2m^2 = 2 \implies m^2 = 1$$

This gives two possible slopes:

• $$m_1 = 1$$
• $$m_2 = -1$$

Step 5: Calculate the Required Value

Substitute the values of $$m_1^2$$ and $$m_2^2$$ into the target expression:

$$4(m_1^2 + m_2^2) = 4(1 + 1) = 4(2) = 8$$

Given parabola

$$y^2=16x$$

Comparing with

$$y^2=4ax$$

we get

$$a=4$$

Hence the focus is

$$S=(4,0)$$

Now find the tangent to the parabola at

$$P(16,16)$$

The tangent to $$y^2=16x$$ at $$(x_1,y_1)$$ is $$yy_1=8(x+x_1)$$

Substituting

$$P=(16,16)$$

$$16y=8(x+16)$$

$$x-2y+16=0$$

Therefore the normal at $$P$$ is $$y+2x-48=0$$ or $$y=48-2x$$

Let the centre of the circle be $$O(h,k)$$

Since the circle touches the parabola at $$P$$, 

the radius $$OP$$ lies along the normal.

Hence

$$O=(h,48-2h)$$

The circle passes through both $$P$$ and $$S$$

Therefore

$$OP=OS$$

Using distance formula,

$$ (h-16)^2+(48-2h-16)^2=(h-4)^2+(48-2h)^2 $$

$$ (h-16)^2+(32-2h)^2=(h-4)^2+(48-2h)^2 $$

Expanding,

$$h^2-32h+256+4h^2-128h+1024=h^2-8h+16+4h^2-192h+2304$$

$$5h^2-160h+1280=5h^2-200h+2320$$

$$40h=1040$$

$$h=26$$

Hence

$$k=48-2(26)=-4$$

Therefore the centre is

$$O=(26,-4)$$

Radius squared is

$$r^2=(26-4)^2+(-4)^2$$

$$=22^2+16$$

$$=500$$

Equation of the circle is

$$ (x-26)^2+(y+4)^2=500 $$

To find intersection with x-axis, put

$$y=0$$

$$ (x-26)^2+16=500 $$

$$ (x-26)^2=484 $$

$$x=26\pm22$$

Thus the two intersections are $$x=4,\ 48$$

Since $$S=(4,0)$$ one intersection is the focus itself.

Hence

$$Q=(48,0)$$

Now coordinates are

$$P=(16,16)$$

$$S=(4,0)$$

$$Q=(48,0)$$

The base

$$SQ=48-4=44$$

and the perpendicular height from $$P$$ to the x-axis is $$16$$

Therefore,

$$\text{Area of }\triangle PQS=\frac12\times44\times16$$

$$=352$$

Hence

$$\frac{\text{Area of }\triangle PQS}{11}=\frac{352}{11}$$

$$=32$$

Final Answer :

$$32$$