The tangent to the curve x y=25 at any point on it cuts the coordinate axes at A and B , then the area of the OAB is

Step-by-step solution

Given, curve x y=25 (i) On differentiating Eq. (i), w.r.t. ' x ', we get x d y d x +y=0 x d y d x =-y dy dx =- y x ( dy dx )_ at ( x ₁, y ₁ ) = - y ₁ x ₁ The equation of tangent at the point (x₁, y₁ ) is given by y-y₁= ( d y d x )_ a t (x₁, y₁ ) y-y₁=- y₁ x₁ (x-x₁ ) x₁ y-x₁ y₁=-x₁+x₁ y₁ x₁+y₁=2 x₁ y₁ On dividing throughout by 2 x₁ y₁ , we get x 2 x₁ + y 2 y₁ =1 Since, (x₁, y₁ ) satisfy the curve x y=25 [from Eq. (iii)]