Question

The tangent drawn at any point of the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ meets the axes in A and B. Prove that $OA+OB$ is a constant.

Answer 1

Lets differentiate $\sqrt{x}+\sqrt{y}=\sqrt{a}....\left(1\right)$ wrt to $x$.

$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$

$⟹\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}$..........$\left(2\right)$

Lets assume that the tangent has the equation in this form,

$\frac{x}{OA}+\frac{y}{OB}=1$

$⟹$ slope $=-\frac{OB}{OA}$

From $\left(2\right)$,

$-\frac{\sqrt{y}}{\sqrt{x}}=-\frac{OB}{OA}$

$⟹\sqrt{y}=c\times OB$ & $\sqrt{x}=c\times OA$ where $c$ is a constant.

$\therefore OA+OB=c(\sqrt{x}+\sqrt{y})$

$⟹OA+OB=c\sqrt{a}$........$\left(3\right)$ from $\left(1\right)$

$⟹OA+OB=$ constant.

Hence Proved.