Question

The equation of 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. $Noneofthese$

Answer 1

Answer: (B)

$\frac{x}{\sqrt{x_1}}+\frac{y}{\sqrt{y_1}}=\sqrt{a}$

We have,

$\sqrt{x}+\sqrt{y}=\sqrt{a}$

Differentiation this equation with respect to $x$ and we get,

$\frac{d}{dx}(\sqrt{x}+\sqrt{y})=\frac{d}{dx}\sqrt{a}$

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

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

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

At point $(x_1,y_1)$

${\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=-\frac{\sqrt{y_1}}{\sqrt{x_1}}$

We know that,

Equation of tangent.

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

At point $(x_1,y_1)$

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

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

$y\sqrt{x_1}-y_1\sqrt{x_1}=-x\sqrt{y_1}+x_1\sqrt{y_1}$

$y\sqrt{x_1}+x\sqrt{y_1}=y_1\sqrt{x_1}+x_1\sqrt{y_1}$

On divide $\sqrt{x_1{y}_1}$ both side and we get,

$\frac{y\sqrt{x_1}+x\sqrt{y_1}}{\sqrt{x_1}\sqrt{y_1}}=\frac{y_1\sqrt{x_1}+x_1\sqrt{y_1}}{\sqrt{x_1}\sqrt{y_1}}$

$\frac{y}{\sqrt{y_1}}+\frac{x}{\sqrt{x_1}}=\sqrt{y_1}+\sqrt{x_1}$

$\frac{x}{\sqrt{x_1}}+\frac{y}{\sqrt{y_1}}=\sqrt{a}$

Hence, this is the answer.