Question

The point on the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ the normal at which is parallel to the $x-$axis is

• A. $(0,0)$
• B. $(0,a)$
• C. $(a,0)$
• D. $(a,a)$

Answer 1

The correct option is B

$(0,a)$

The explanation for the correct answer.

Solve for the value of required value.

Given: Equation of the curve is $\sqrt{x}+\sqrt{y}=\sqrt{a}$…………….$\left(1\right)$

Differentiate with respect to $x$,

$\left(\frac{1}{2\sqrt{x}}\right)+\left(\frac{1}{2\sqrt{y}}\right)y\text{'}=0$ $\left[\frac{d{a}^n}{dx}=n{a}^{n-1}\frac{da}{dx}\right]$

$\Rightarrow$ $\left(\frac{1}{2\sqrt{y}}\right)y\text{'}=-\frac{1}{2\sqrt{x}}$

$\Rightarrow$ $y\text{'}=–\sqrt{\frac{y}{x}}$

We know that the slope $\left(m\right)$ of normal is given by, $–\frac{1}{y\text{'}}$

$$\begin{gathered} m=-\frac{1}{y\text{'}} \\ \Rightarrow m=\sqrt{\frac{x}{y}} \end{gathered}$$

Since, the normal is parallel to $x$ axis.

Thus, $m=0$

$$\begin{gathered} \sqrt{\frac{x}{y}}=0 \\ \Rightarrow x=0 \end{gathered}$$

Putting the value of $x$ in equation $\left(1\right)$

$$\begin{gathered} \sqrt{x}+\sqrt{y}=\sqrt{a} \\ \Rightarrow 0+\sqrt{y}=\sqrt{a} \\ \Rightarrow y=a \end{gathered}$$

$y=a,x=0$

Therefore, the required point is $(0,a)$.

Hence, option(B) is the required answer.