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

Step-by-step solution

Given curve, x^2 a^2 + y^2 b^2 =1 Differentiate w.r.t x , we get 2 x a^2 + 2 y y^ b^2 =0 y^ = -b^2 x y a^2 At ( a 2 , b 2 ), y^ = -b^2 ( a 2 ) b 2 (a^2 ) =- b a So, slope of tangent, m₁=- b a and slope of normal, m₂= a b Equation of tangent at ( a 2 , b 2 ) is (y- b 2 )= -b a (x- a 2 ) And equation of normal at ( a 2 , b 2 ) is (y- b 2 )= a b (x- a 2 ) . Now, Area of = 1 2 base height = 1 2 ( a^2+b^2 2 a ) b 2 = b 4 a (a^2+b^2 ) .