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Q. 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

AP EAMCETAP EAMCET 2019

Solution:

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