Question

The sum of intercept made on coordinate axes made by tangent to curve √x +√y =√a is

Answer 1

The sum of intercept made on coordinate axes made by tangent to curve √x + √y = √a

solution : Let (x₁, y₁) is the point on the curve.

so, √x₁ + √y₁ = √a ..........(1)

now differentiating curve with respect to x.

⇒-1/2√x - 1/2√y dy/dx = 0

⇒-1/2√x = 1/2√y dy/dx

⇒dy/dx = -√(y/x)

so slope of tangent at (x₁, y₁), m = -√(y₁/x₁)

now equation of tangent is given by,

(y - y₁) = m(x - x₁)

⇒y - y₁ = -√(y₁/x₁)(x - x₁)

⇒√x₁(y - y₁) = -√y₁ (x - x₁)

⇒y√x₁ - y₁√x₁ = -x√y₁ + x₁√y₁

⇒y√x₁ + x√y₁ = y₁√x₁ + x₁√y₁ =√x₁y₁(√y₁ + √x₁)

⇒y/√y₁(√y₁ + √x₁) + x/√x₁(√y₁ + √x₁) = 1

now sum of intercept made on coordinate axes

= √y₁(√y₁ + √x₁) + √x₁(√y₁ + √x₁)

= (√y₁ + √x₁)(√y₁ + √x₁)

from equations (1) we get

= √a × √a

= a

Therefore the sum of intercept made on coordinate axes made by tangent to curve √x +√y =√a is a.