Question

The sum of the intercepts made on the axes of coordinates by any tangent to the curve $\sqrt{x}+\sqrt{y}=2$ is equal to :

• A. $4$
• B. $2$
• C. $8$
• D. None of these

Answer 1

The correct option is A $4$

Let P $(x_1,y_1)$ be any point on the curve $\sqrt{x}+\sqrt{y}=2$

For the curve, $\sqrt{x}+\sqrt{y}=2$

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

=> $\frac{dy}{dx}=-\sqrt{\frac{y}{x}}$

=> $\frac{dy}{dx}atP=-\sqrt{\frac{y_1}{x_1}}$

=> Equation of tangent at P to the curve is

$y-y_1=-\sqrt{\frac{y_1}{x_1}}(x-x_1)$

plugging y = 0,

x-intercept is $\sqrt{x_1{y}_1}+x_1=\sqrt{x_1}(\sqrt{y_1}+\sqrt{x_1})$

plugging x = 0,

y-intercept is $\sqrt{x_1{y}_1}+y_1=\sqrt{y_1}(\sqrt{x_1}+\sqrt{y_1})$

=> sum of the intercepts = $\sqrt{x_1}(\sqrt{y_1}+\sqrt{x_1})+\sqrt{y_1}(\sqrt{x_1}+\sqrt{y_1})$

= $(\sqrt{x_1}+\sqrt{y_1})\times (\sqrt{y_1}+\sqrt{x_1})$

= $2\times 2=4$