Question

The equation of the tangent to the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ at the point $(x_1,y_1)$ is -

• A. $\frac{x}{\sqrt{x_1}}+\frac{y}{\sqrt{y_1}}=\frac{1}{\sqrt{a}}$
• B. $\frac{x}{\sqrt{x_1}}+\frac{y}{\sqrt{y_1}}=\sqrt{a}$
• C. $x\sqrt{x_1}+y\sqrt{y_1}=\sqrt{a}$
• D. None of these

Answer 1

The correct option is B $\frac{x}{\sqrt{x_1}}+\frac{y}{\sqrt{y_1}}=\sqrt{a}$

$\sqrt{x}+\sqrt{y}=\sqrt{a}$ at point $(x_1,y_1)$
$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$
$\frac{dy}{dx}=-\sqrt{\frac{y}{x}}\Rightarrow {\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=-\frac{\sqrt{y_1}}{\sqrt{x_1}}$
Thus equation of tangent is given by, $y-y_1=-\frac{\sqrt{y_1}}{\sqrt{x_1}}(x-x_1)$
$y\sqrt{x_1}-y_1\sqrt{x_1}=-x\sqrt{y_1}+x_1\sqrt{y_1}$
$x\sqrt{y_1}+y\sqrt{x_1}=\sqrt{x_1}\sqrt{y_1}\left(\sqrt{a}\right)$
$\frac{x}{\sqrt{x_1}}+\frac{y}{\sqrt{y_1}}=\sqrt{a}$
Hence, option 'B' is correct.