If the tangent at a point P on the parabola y 2 = 3 x is parallel to the line x + 2 y = 1 and the tangents at the points Q and R on the ellipse x 2 4 + y 2 1 = 1 are perpendicular to the line x - y = 2 , then the area of the triangle P Q R is:

Step-by-step solution

Given equation of parabola is y 2 = 3 x ⇒ 2 y d y d x = 3 ⇒ d y d x slope   of   tangent   at   P = 3 2 y Tangent at P x 1 , y 1 is parallel to the line x + 2 y = 1 , so slope of tangent is m = - 1 2 ⇒ 3 2 y 1 = - 1 2 ⇒ y 1 = - 3 So, coordinates of P ≡ 3 , - 3 . Given equation of ellipse is x 2 4 + y 2 1 = 1 ⇒ x 2 + 4 y 2 = 4 ⇒ 2 x + 8 y d y d x = 0 ⇒ d y d x slope   of   tangent = - x 4 y Also, tangent to ellipse re perpendicular to line