Question

The equation of the tangent to the curve $\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$ at the point $(x_1,y_1)$ is $\frac{x}{\sqrt{a{x}_1}}+\frac{y}{\sqrt{b{y}_1}}=k$. Then, the value of $k$ is

• A. $2$
• B. $1$
• C. $3$
• D. $7$
• E. $\sqrt{2}$

Answer 1

Answer: (B)

$1$

Given curve is
$\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$ ... (i)
On differentiating w.r.t. to $x$, we get
$\frac{1}{\sqrt{a}}\cdot \frac{1}{2\sqrt{x}}+\frac{1}{\sqrt{b}}\cdot \frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{\sqrt{b}\sqrt{y}}{\sqrt{a}\sqrt{x}}\Rightarrow {\left[\frac{dy}{dx}\right]}_{(x_1,y_1)}=\frac{-\sqrt{b}\sqrt{y_1}}{\sqrt{a}\sqrt{x_1}}$
Equation of tangent passing through the point $(x_1,y_1)$ is
$(y-y_1)=\frac{-\sqrt{b}\sqrt{y_1}}{\sqrt{a}\sqrt{x_1}}(x-x_1)$
$\frac{y}{\sqrt{b{y}_1}}-\frac{y_1}{\sqrt{b{y}_1}}=-\frac{x}{\sqrt{a{x}_1}}+\frac{x_1}{\sqrt{a{x}_1}}$
$\Rightarrow \frac{x}{\sqrt{a{x}_1}}+\frac{y}{\sqrt{b{y}_1}}=\frac{x_1}{\sqrt{a{x}_1}}+\frac{y_1}{\sqrt{b{y}_1}}=\sqrt{\frac{x_1}{a}}+\sqrt{\frac{y_1}{b}}$
$\Rightarrow \frac{x}{\sqrt{a{x}_1}}+\frac{y}{\sqrt{b{y}_1}}=1$ [from Eq. (i)]
$[\because at(x_1,y_1),\sqrt{\frac{x_1}{a}}+\sqrt{\frac{y_1}{b}}=1]$
But $\frac{x}{\sqrt{a{x}_1}}+\frac{y}{\sqrt{b{y}_1}}=k$ (given)
Therefore, $k=1$