Question

Number of normals at the points on the curve $y=\frac{x}{(1-x^2)}$ where the tangents makes an angle of $\frac{\pi }{4}$ with x-axis

Answer 1

Given eqn of curve is
$y=\frac{x}{1-x^2}$
Also, given slope of tangent to curve is $\tan \frac{\pi }{4}=1$
$\frac{dy}{dx}=\frac{1.(1-x^2)-(-2x)x}{(1-x^2{)}^2}$
$\Rightarrow 1=\frac{1+x^2}{(1-x^2{)}^2}$
$\therefore (1+x^2)=(1-x^2{)}^2$
$x^4-3x^2=0$
$\Rightarrow x=0,-\sqrt{3},\sqrt{3}$
At $x=0\Rightarrow y=0$
At $x=\sqrt{3}\Rightarrow y=\frac{-\sqrt{3}}{2}$
At $x=-\sqrt{3}\Rightarrow y=\frac{\sqrt{3}}{2}$
Hence the points are : $(0,0),(-\sqrt{3},\frac{\sqrt{3}}{2})and(\sqrt{3},-\frac{\sqrt{3}}{2})$.
Eqn of normal at (0,0) is
$y-0=-1(x-0)$
$x+y=0$
Eqn of normal at $(\sqrt{3},-\frac{\sqrt{3}}{2})$ is
$y+\frac{\sqrt{3}}{2}=-1(x-\sqrt{3})$
$2y+2x=\sqrt{3}$
Eqn of normal at $(-\sqrt{3},\frac{\sqrt{3}}{2})$ is
$y-\frac{\sqrt{3}}{2}=-1(x+\sqrt{3})$
$2y+2x=-\sqrt{3}$