How to find the angle between the two tangents on a parabola 
Two tangents are drawn from the point $(-2,-1)$ to the parabola $y^2=4x$.
  If $\alpha$ is the angle between them then find the value of $\tan \alpha$ .

My try:
Eqn of tangent $1,T_1$ say at the point $(x_1,y_1)$ is $yy_1=2a(x+x_1)$,
Eqn of tangent $2,T_2$ say at the point $(x_2,y_2)$ is $yy_2=2a(x+x_2)$
Since they both pass through $(-2,-1)$ hence we have $y_1=2(2-x_1),y_2=2(2-x_2)$.
Angle between two tangents is $\tan \alpha=\vert\dfrac{m_1-m_2}{1+m_1m_2}\lvert$
Now $m_1=\dfrac{y_1+1}{x_1+2},m_2=\dfrac{y_2+1}{x_2+1}$
Replacing $y_1,y_2$ is not giving the required angle.
How to do it?Please help.
 A: Any point on $y^2=4x$ P$(t^2,2t)$
$$\dfrac{dy}{dx}_{(\text{ at }P)}=\dfrac4{2y}_{(\text{ at }P)}=\dfrac1t$$
$$\implies\dfrac1t=\dfrac{2t+1}{t^2+2}\iff t^2+t-2=0$$
If $t_1.t_2$ are the roots of the above equation, WLOG  $t_1=1,t_2=-2$
So, $\tan\alpha=\left|\dfrac{\dfrac1{t_1}-\dfrac1{t_2}}{1+\dfrac1{t_1}\cdot\dfrac1{t_2}}\right|=\left|\dfrac{t_2-t_1}{t_1t_2+1}\right|$
A: Taking the derivative wrt $x$, $y^2=4x$ becomes $2yy'= 4$, or $y' = \dfrac{2}{y}$.
So, at the point $(\frac 14t^2,t)$ on the parabola, the slope of the tangent is 
$m = \dfrac 2t$.
So the equation of the tangent line through the point $(\frac 14t^2,t)$
is $y-t = \frac 2t(x-\frac 14t^2)$.
For what values of $t$ does this line pass through the point $(-2,-1)$?
\begin{align}
   -1-t &= \frac 2t(-2-\frac 14t^2) \\
   1+t &= \frac 4t + \frac t2\\
   2t + 2t^2 &= 8 + t^2\\
   t^2+2t-8 &= 0 \\
   (t+4)(t-2) &= 0 \\
   t &\in \{2, -4\} \\
   (x,y) &\in \{(1,2), (4,-4)\} \\
   m &\in \left\{1, -\frac 12 \right\}
\end{align}
\begin{align}
   \tan\alpha
   &=\left|\dfrac{m_1-m_2}{1+m_1 m_2}\right| \\
   &=\left|\dfrac{1+\frac 12}{1-\frac 12}\right| \\
   &= 3
\end{align}
$$\alpha \approx 71.57^\circ$$

