JEE Advanced | 2014CircleHard

Question

The common tangents to the circle x2 + y2 = 2 and the parabola y2 = 8x touch the circle at the points P, Q and the parabola at the points R, S. Then the area of the quadrilateral PQRS is

Options

A.3
B.6
C.9
D.15

Solution

       
x2 + y2 = 2
Equation of tangent to circle x2 + y2 = 2 is y = mx + √2
Equation of tangent to y2 = 8x is y = mx +

2(1 + m2) =
m2(1 + m2) = 2
m4 + m2 - 2 = 0
(m2 + 2)(m2 - 1) = 0
m = + 1
y = x + 2
y = -x - 2
Area = × 2 = 6 + 9 = 15 sq. unit

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