Question

The number of tangents to the curve $y^2(x-a)=x^2(x+a)(a>0)$ that are parallel to the X-axis is

• A. infinitely many
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (C)

Given, curve is
$y^2(x-a)=x^2(x+a)$
Apply log on both sides
$\log \{y^2(x-a\}=\log \left\{x^2(x+a)\right\}$
$\Rightarrow 2\log y+\log (x-a)=2\log x+\log (x+a)$
Differentiating both sides w.r.t. $x$, we get
$\Rightarrow \frac{2}{y}\cdot \frac{dy}{dx}+\frac{1}{x-a}$
$=\frac{2}{x}+\frac{1}{x+a}$
$\Rightarrow \frac{2}{y}\frac{dy}{dx}$
$=\frac{2}{x}+\frac{1}{x+a}-\frac{1}{x-a}$
$=\frac{2(x+a)(x-a)+x(x-a)-x(x+a)}{x(x+a)(x-a)}$
$=\frac{2\left(x^2-a^2\right)+x^2-ax-x^2-ax}{x(x-a)(x+a)}$
$\Rightarrow \frac{2}{y}\frac{dy}{dx}$
$=\frac{2x^2-2a^2-2ax}{x(x-a)(x+a)}$
$\Rightarrow \frac{dy}{dx}=\frac{x^2-a^2-ax}{x(x-a)(x+a)}\cdot x\cdot \sqrt{\frac{x+a}{x-a}}$
At $\frac{dy}{dx}=0$
$\therefore x^2-a^2-ax=0$
Here, $p=a^2+4a^2>0$
So. it has two real roots.