Question

The number of tangents to the curve $x^{\frac{3}{2}}+y^{\frac{3}{2}}=2a^{\frac{3}{2}},a>0$, which are equally inclined to the axes, is

• A. $2$
• B. $1$
• C. $0$
• D. $4$

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Given: curve $x^{\frac{3}{2}}+y^{\frac{3}{2}}=2a^{\frac{3}{2}}...\left(i\right)$

Differentiating both sides with respect to $x$, we get

$$\begin{gathered} \frac{3}{2}\sqrt{x}+\frac{3}{2}\sqrt{y}\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-\sqrt{x}}{\sqrt{y}} \end{gathered}$$

Tangent is equally inclined to the axes.

$\therefore \frac{dy}{dx}=\pm 1$

So,$\frac{-\sqrt{x}}{\sqrt{y}}=\pm 1$

$$\begin{gathered} \Rightarrow -\sqrt{x}=\pm \sqrt{y} \\ \Rightarrow x=y \end{gathered}$$

Put$x=y$ in equation $\left(i\right)$, we get

$$\begin{gathered} x^{\frac{3}{2}}+x^{\frac{3}{2}}=2a^{\frac{3}{2}} \\ \Rightarrow 2x^{\frac{3}{2}}=2a^{\frac{3}{2}} \\ \Rightarrow x=a \end{gathered}$$

$\therefore x=y=a$

As, the tangent touches the curve at $(a,a)$.Thus we get only one tangent, satisfying that criteria.

Therefore, the number of tangents possible $=1$

Hence, option B is the answer.