The area enclosed by the closed curve C given by the differential equation d y d x + x + a y - 2 = 0 , y 1 = 0 is 4 π . Let P and Q be the points of intersection of the curve C and the y -axis. If normals at P and Q on the curve C intersect x -axis at points R and S respectively, then the length of the line segment R S is

Step-by-step solution

Given: d y d x + x + a y - 2 = 0 ⇒ d y d x = x + a 2 - y ⇒ 2 - y   d y = x + a   d x ⇒ 2 y - y 2 2 = x 2 2 + a x + c Put x = 1 , then we get y 1 = 0 . a + c = - 1 2 Now, 4 y - y 2 = x 2 + 2 a x + 2 c ⇒ x 2 + y 2 + 2 a x - 4 y + 2 c = 0 ⇒ x 2 + y 2 + 2 a x - 4 y - 1 - 2 a = 0 This is a circle having radius r = a 2 + 4 + 1 + 2 a = a + 1 2 + 4 Now, π r 2 = 4 π r 2 = 4 ⇒ a + 1 2 + 4 = 4 ⇒ a + 1 2 = 0 So, a = - 1 Hence, circle is x 2 + y 2 - 2 x - 4 y + 1 = 0