Question

The area of cyclic quadrilateral OABC is?

• A. $24\sqrt{3}$
• B. $48\sqrt{2}$
• C. $12\sqrt{6}$
• D. $18\sqrt{5}$

Answer 1

The correct option is A $24\sqrt{3}$

Solving given parabolas, we have

$-8(x-a)=4x$

$\Rightarrow x=\frac{2a}{3}$

$\Rightarrow$ Points of intersection are $(\frac{2a}{3},\pm \sqrt{\frac{8a}{3}})$

Now $OABC$ is concyclic.

Fig.3.133

Hence, $\angle OAB$ must be right angle.

$\Rightarrow$ Slope of $OA\times$ Slope of $AB=-1$

$\Rightarrow \frac{\sqrt{\frac{8a}{3}}}{\frac{2a}{3}}\times \frac{\sqrt{\frac{8a}{3}}}{a-\frac{2a}{3}}=-1$

$\Rightarrow a=12$

$\Rightarrow$ Coordinates of $A$ and $B$ are $(8,4\sqrt{2})$ and $(8,-4\sqrt{2})$ respectively

$\Rightarrow$ Length of common chord $=8\sqrt{2}$

$\text{Area of quadrilateral}=\frac{1}{2}OB\times AC$

$=\frac{1}{2}\times 12\times 8\sqrt{2}$

$=48\sqrt{2}$

Tangent to parabola $y^2=4x$ at point $(8,4\sqrt{2})$ is $4\sqrt{2}y=2(x+8)$ or $x-2\sqrt{2}y+8=0$ which meets the $x$ -axis at

$D(-8,0)$

Tangent to parabola $y^2=-8(x-12)$ at point $(8,4\sqrt{2})$ is

$4\sqrt{2}y=-4(x+8)+96$ or $x+\sqrt{2}y-16=0,$ which meets the $x$ -axis at $E(16,0)$

Hence, area of quadrilateral $DAEC=\frac{1}{2}DE\times AC$

$=\frac{1}{2}\times 24\times 8\sqrt{2}$

$=96\sqrt{2}$