The correct option is
A 24√3 Solving given parabolas, we have
−8(x−a)=4x
⇒x=2a3
⇒ Points of intersection are (2a3,±√8a3)
Now OABC is concyclic.
Fig.3.133
Hence, ∠OAB must be right angle.
⇒ Slope of OA× Slope of AB=−1
⇒√8a32a3×√8a3a−2a3=−1
⇒a=12
⇒ Coordinates of A and B are (8,4√2) and (8,−4√2) respectively
⇒ Length of common chord =8√2
Area of quadrilateral =12OB×AC
=12×12×8√2
=48√2
Tangent to parabola y2=4x at point (8,4√2) is 4√2y=2(x+8) or x−2√2y+8=0 which meets the x -axis at
D(−8,0)
Tangent to parabola y2=−8(x−12) at point (8,4√2) is
4√2y=−4(x+8)+96 or x+√2y−16=0, which meets the x -axis at E(16,0)
Hence, area of quadrilateral DAEC=12DE×AC
=12×24×8√2
=96√2