How to get the Normal line? My book proposed to me to find the the Normal lines to the curve that pass through the origin.The answer must be the intersection points between them.
The curve: $\dfrac{2}{1+x^2}$
My first idea is to get the derivative: $f'(x) = $ $\frac{-4x}{x^4 + 2x^2 + 1}$, then get the perpendicular of $f'(x)$, resulting in 
$M = \dfrac{-1}{f'(x)}$ = $\dfrac{x^4 + 2x^2 + 1}{4x}$.
The segment that pass through the origin with the angular coeficient $M$ is: 
$g(x) = M * (x - x0) + y0$; to $x0$ = 0 and $y0$ = 0.
So, $g(x) = \dfrac{x^4 + 2x^2 + 1}{4}$
Finally to get the common points between $g(x)$ and $f(x)$, I have the equality: $f(x) = g(x)$.
$\dfrac{x^4 + 2x^2 + 1}{4}$ = $\dfrac{2}{x^2 + 1}$
Resulting in two real solutions: $x = -1$ and $x = 1$ and another four complex roots. If you look carefully, we can see a "hidden" solution at x = 0.
Is it the right way to procced to get the points !?

If $\displaystyle x=\tan\theta, y=\frac2{1+x^2}=2\cos^2\theta$
So, $\displaystyle\frac{dy}{dx}=-\frac{4x}{(1+x^2)^2}=-4\tan\theta\cos^4\theta=-4\sin\theta\cos^3\theta$
So, the equation of the normal at $(\tan\theta,2\cos^2\theta)$
$$\frac{y-2\cos^2\theta}{x-\tan\theta}=-\frac1{-4\sin\theta\cos^3\theta}$$
$$\implies(y-2\cos^2\theta)4\sin\theta\cos^4\theta=x-\sin\theta \ \ \ \ (1)$$
Now as the normal passes through the origin, set $x=y=0$ to find the required values of $\theta$ 
$$\implies(0-2\cos^2\theta)4\sin\theta\cos^4\theta=0-\sin\theta\iff\sin\theta(8\cos^6\theta-1)=0$$
If $\displaystyle\sin\theta=0\implies\cos^2\theta=1, (1)$ becomes $\displaystyle y-2=x$
If $\displaystyle8\cos^6\theta-1=0,$ the only real value of $\displaystyle\cos^2\theta$ is $\displaystyle\frac12\implies\sin\theta=\pm\frac1{\sqrt2}$
Put the values of  $\displaystyle\cos^2\theta,\sin\theta$ in $(1)$
