Let P be the point on the parabola y = x2 such that the slope of the tangent to the parabola at the point P is 4 . Let Q be the point in the first quadrant lying on the circle x2 + y2 = 2 such that the slope of the tangent to the circle at the point Q is -1 . Let R be the point in the first quadrant lying on the ellipse x2 + 4y2 = 8 such that the slope of the tangent to the ellipse at the point R is - 12 . Then the radius of the circle passing through the points P, Q and R is
- 10
- 5
- 52
- 2 5
Answer
(3) 52
Solution
Find the three points from the slope conditions.
Parabola: dydx = 2x = 4 P(2, 4).
Circle: slope - xy = -1 x = y, 2x2 = 2 Q(1, 1).
Ellipse: slope - x4y = - 12 x = 2y, 8y2 = 8 R(2, 1).
PR is vertical (x = 2) and QR is horizontal (y = 1), so PRQ = 90 and PQ is a diameter.
PQ = 1 + 9 = 10, radius = 102 = 52
Answer: (C)