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 the radius of the circle passing through the points P, Q and R is
C. 5 2
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 = 8, differentiating with respect to x gives 2x + 8y dy dx = 0 dy dx = - x 4y . Given - x 4y = - 1 2 x = 2y. Since R lies in the first quadrant, substituting x = 2y in x^2 + 4y^2 = 8 gives 4y^2 + 4y^2 = 8 8y^2 = 8 y = 1, x = 2. Thus, the coordinates of R are (2, 1). The points are P(2, 4), Q(1, 1), and R(2, 1). The line segment PR lies on the vertical line x = 2, and the line segment QR lies on the horizontal line y = 1. Since PR is perpendicular to QR, the angle PRQ = 90^ . Therefore, the triangle PQR is a right-angled triangle with the hypotenuse PQ. The circumcircle of PQR has PQ as its diameter. The length of the diameter is PQ = (2 - 1)^2 + (4 - 1)^2 = 1^2 + 3^2 = 10 . The radius of the circle is PQ 2 = 10 2 = 5 2 . Answer: 5 2
Related: Mathematics — Application of Derivatives · All PYQ Banks