The area (in sq. units) of triangle formed by the tangent and normal at a point a 2 , b 2 to the curve x 2 a 2 + y 2 b 2 = 1 and X -axis is

Step-by-step solution

We need to find the area of ∆ P M N a r ∆ P M N = 1 2 × Length of normal × Length of tangent Now, x 2 a 2 + y 2 b 2 = 1 ⇒ d y d x = - b x a a 2 - x 2 Now, length of normal is given by = y 1 + d y d x 2 = b 2 1 + - b x a a 2 - x 2 2 At x = a 2 = b 2 a a 2 + b 2 Now, length of tangent is given by y 1 + d x d y 2 = a 2 + b 2 2 Therefore, area of triangle is = 1 2 a 2 + b 2 2 b 2 a a 2 + b 2 = b 4 a a 2 + b 2