`y^(2)=4xandy^(2)=-8(x-a)` intersect at points A and C. Points O(0,0), A,B (a,0), and c are concyclic.
The area of cyclic quadrilateral OABC is

Options

A. `24sqrt(3)`
B. `48sqrt(2)`
C. `12sqrt(6)`
D. `18sqrt(5)`

Answer

B

Explanation

(2)
Solving the fiven parabolas , we have
-8(x-1)=4x
`orx=(2a)/(3)`
Therefore, the intersection are `(2a//3,pmsqrt(8a//3))`.
Now, OABC is cyclic quadrilateral.

Hence, `angleOAB` must be a right angle. So,
Slope of `OAxx` Slope of AB=-1
`or(sqrt(8a//3))/(2a//3)xx(sqrt(8a//3))/(a-(2a//3))=-1`
`ora=12`
Therefore, the coordinates of A and B are `(8,4sqrt(2))and(8,-4sqrt(2))`, respectively. So,
Length of common chord `=8sqrt(2)`
Area of quadrilateral `=(1)/(2)OBxxAC`
`=(1)/(2)xx12xx8sqrt(2)`
`48sqrt(2)`
Tangent to the parabola `y^(2)=4xat(8,4sqrt(2))" is "4sqrt(2)y=2(x+8)orx-2sqrt(2)y+8=0`, which meets the x-axis at D(-8,0).
Tangent to the parabola `y^(2)=-8(x-12)at(8,4sqrt(2))" is "4sqrt(2)y=-4(x+8)+96orx+sqrt(2)y=16=0`, which meets the x-axis at E(16,0). Hence,
Area of quadrilateral `DAEC=(1)/(2)DExxAC`
`(1)/(2)xx24xx8sqrt(2)`
`=96sqrt(2)`