Question

The sum of intercepts on coordinate axes made by tangent to the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ is

• A. $a$
• B. $2a$
• C. $2\sqrt{a}$
• D. None of these

Answer 1

The correct option is A

$a$

Explanation for the correct option:

Find the required sum.

Given curve: $\sqrt{x}+\sqrt{y}=\sqrt{a}$

On differentiating w.r.t $x$, we get

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

$\Rightarrow \frac{dx}{dy}=-\sqrt{\frac{y}{x}}$ ( slope of tangent)

To find points,

Let $x=c$

$y={(\sqrt{a}-\sqrt{c})}^2$

So, equation of tangent ,

$$\begin{gathered} Y-{\left(\sqrt{a}-\sqrt{c}\right)}^2=-\sqrt{\frac{y}{x}}\left(X-c\right) \\ \Rightarrow \frac{Y-{\left(\sqrt{a}-\sqrt{c}\right)}^2}{\left(x-c\right)}=-\sqrt{\frac{y}{x}} \\ \Rightarrow \frac{Y-{\left(\sqrt{a}-\sqrt{c}\right)}^2}{\left(X-c\right)}=-\sqrt{\frac{{\left(\sqrt{a}-\sqrt{c}\right)}^2}{c}}[\because x=c\&y={\left(\sqrt{a}-\sqrt{c}\right)}^2] \end{gathered}$$

$\Rightarrow \frac{Y-{\left(\sqrt{a}-\sqrt{c}\right)}^2}{\left(X-c\right)}=\frac{\sqrt{c}-\sqrt{a}}{c}$ ….(i)

For Intercept of $x$,

put$Y=0$ in equation (i) we get,

$X=\sqrt{ac}$ ….(ii)
For intercept of $y$,

put$X=0$ in equation (i) we get,

$Y=a-\sqrt{ac}$ …(iii)

Sum of both Intercepts (ii)+(iii)

$\sqrt{ac}+a-\sqrt{ac}=a$

Hence, option (A) is the correct answer.