Question

$y^2=4x$ and $y^2=-8(x-a)$ intersect at point $A$ and $C$. Points $O(0,0),A,B(a,0),C$ are concyclic.
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 B $48\sqrt{2}$

Solving the given parabolas, we have
$-8(x-a)=4x$
$\Rightarrow x=\frac{2a}{3}$
Therefore, the points of intersection are $(\frac{2a}{3},\pm \sqrt{\frac{8a}{3}})$
Now, OABC is noncyclic.
Hence, $\angle OAB$ must be a right angle. So,
Slope of $\text{OA}\times \text{Slope of AB}=-1$
$\Rightarrow \frac{\sqrt{\frac{8a}{3}}}{\frac{2a}{3}}\times \frac{\sqrt{\frac{8a}{3}}}{a-\left(\frac{2a}{3}\right)}=-1$
$\Rightarrow a=12$
Therefore, the coordinates of $A$ and $C$ are $(8,4\sqrt{2})$ and $(8,-4\sqrt{2})$ respectively.
So, Length of common chord $=8\sqrt{2}$

Area of quadrilateral $=\frac{1}{2}OB\times AC$
$=\frac{1}{2}\times 12\times 8\sqrt{2}$
$=48\sqrt{2}$ sq. units