Question

The number of tangent(s) to the curve $x^{3/2}+y^{3/2}=2a^{3/2},a>0,$ which are equally inclined to the axes is/are

• A. $2$
• B. $1$
• C. $0$
• D. $4$

Answer 1

The correct option is B $1$

Given curve is $x^{3/2}+y^{3/2}=2a^{3/2}\cdots \left(i\right)$
Differentiate w.r.t. $x$, we get
$\frac{3}{2}\sqrt{x}+\frac{3}{2}\sqrt{y}\frac{dy}{dx}=0$
or $\frac{dy}{dx}=-\frac{\sqrt{x}}{\sqrt{y}}$
Since the tangent is equally inclined to the axes
$\therefore \frac{dy}{dx}=\pm 1$
$\therefore -\frac{\sqrt{x}}{\sqrt{y}}=\pm 1\Rightarrow -\frac{\sqrt{x}}{\sqrt{y}}=-1[\because \sqrt{x}>0,\sqrt{y}>0]$
$\Rightarrow \sqrt{x}=\sqrt{y}$
Putting $\sqrt{y}=\sqrt{x}$ in $\left(i\right)$, we get
$2x^{3/2}=2a^{3/2}\Rightarrow x^3=a^3$
$\therefore x=a$ and so $y=a$.