Let P be the point on the parabola y = x^2 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 x^2 + y^2 = 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 x^2 + 4y^2 = 8 such that the slope of the tangent to the ellipse at the point R is - 1 2 . Then

Step-by-step solution

For the point P on the parabola y = x^2 , the slope of the tangent is dy dx = 2x . Given 2x = 4 x = 2 . Substituting x = 2 in y = x^2 , we get y = 4 . Thus, the coordinates of P are (2, 4) . For the point Q on the circle x^2 + y^2 = 2 , differentiating with respect to x gives 2x + 2y dy dx = 0 dy dx = - x y . Given - x y = -1 x = y . Since Q lies in the first quadrant, substituting x = y in x^2 + y^2 = 2 gives 2x^2 = 2 x = 1, y = 1 . Thus, the coordinates of Q are (1, 1) . For the point R on the ellipse x^2 + 4y^2