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.
Tangents to parabola $y^2=4x$ at $A$ and $C$ intersect at point $D$ and tangents to parabola $y^2=-8(x-a)$ intersect at point $E$, then the area of quadrilateral $DAEC$ is

• A. $96\sqrt{2}$
• B. $48\sqrt{3}$
• C. $54\sqrt{5}$
• D. $36\sqrt{6}$

Answer 1

The correct option is A $96\sqrt{2}$

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 $=\frac{1}{2}OB\times AC$
$=\frac{1}{2}\times 12\times 8\sqrt{2}$
$=48\sqrt{2}$

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

Tangent to the parabola $y^2=-8(x-12)$ at $(8,4\sqrt{2})$ is
$4\sqrt{2}y=-4(x+8)+96$
$\Rightarrow x+\sqrt{2}y-16=0$, which meets the $x-$ axis at $E(16,0)$.
Hence,
Area of quadrilateral $DAEC=\frac{1}{2}DE\times AC=\frac{1}{2}\times 24\times 8\sqrt{2}=96\sqrt{2}$