Range of $\alpha$ If tangents are drawn from external point to the Hyperbola 
Two tangents can be drawn to the different branches of the hyperbola $$\frac{x^2}{1}-\frac{y^2}{4}=1$$ from the point $(\alpha,\alpha^2)$. Then Range of $\alpha$ is

$\bf{My\; Try::}$If Line $y=mx+c$ is tangent to the hyperbola, Then we get equation of
tangent in slope form as $$y=mx\pm \sqrt{a^2m^2-b^2}$$
so here we get $$y=mx\pm \sqrt{m^2-4}$$
Now tangents line passess through $(\alpha,\alpha^2).$ 
So we get $$\alpha^2=m\alpha\pm \sqrt{m^2-4}$$
So $$(\alpha^2-m\alpha)^2=m^2-4\Rightarrow \alpha^4+m^2\alpha^2-2m\alpha^3=m^2-4$$
So $$\alpha^4-2m\alpha^3+m^2\alpha^2-m^2+4=0$$
Now how can i solve after that, Help me
Thanks
 A: Note that
$$(\alpha^2-1)m^2-2\alpha^3m+\alpha^2+4=((\alpha-1)m-(\alpha^2-2\alpha+2))((\alpha+1)m-(\alpha^2+2\alpha+2))$$


*

*If $\alpha\notin\{-1,1\}$ then your equation has two solutions in $m$,
These two solutions are distinct if and only if $\alpha\notin\{-\sqrt{2},\sqrt{2}\}$, and this gives the desired two tangents to the hyperbola. While if $\alpha\in\{-\sqrt{2},\sqrt{2}\}$ then $(\alpha,\alpha^2)$ is on the hyperbola
and there is just one tangent through that point.

*If $\alpha\in\{-1,1\}$ then the equation gives the slop of one tangent, but there is also another vertical tangent through that point to the hyperbola. 


We conlude that the desired range of alpha is  $\alpha\notin\{-\sqrt{2},\sqrt{2}\}$.
A: The equation of tangent at $P(\sec t,2\tan t)$ is $$x\sec t-\dfrac y4\cdot2\tan t=1\iff2x-y\sin t=2\cos t$$
If this has to be pass through  $(\alpha,\alpha^2),$
$$2\alpha-\alpha^2\sin t=2\cos t$$
$$\implies\alpha=\tan\dfrac t2+1,\cot\dfrac t2-1$$
So, $\alpha$ can assume any real value.
