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.
Best Answer
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, $\tan\alpha=\left|\dfrac{\dfrac1{t_1}-\dfrac1{t_2}}{1+\dfrac1{t_1}\cdot\dfrac1{t_2}}\right|=\left|\dfrac{t_2-t_1}{t_1t_2+1}\right|$