Point of intersection of tangents If the distance of two points $P$ and $Q$ from the focus of of a parabola $y^2 =4ax$ are $4$ & $9$  then what is the distance of the point of intersection of tangents at $P$ and $Q$ from the focus.
I assume the points 
$P(a+4\cos\theta, \sin\theta)$ &  $Q (a-4a\cos\theta, \sin\theta)$
These points satisfy the equation of parabola. But  I found no result.
 A: Notice, let the points $P(at_1^2, 2at_1)$ & $Q(at_2^2, 2at_2)$ on the parabola: $y^2=4ax$ 
then the distance of the point $P(at_1^2, 2at_1)$ from the focus $S(a, 0)$ is given as $$PS=\sqrt{(at_1^2-a)^2+(2at_1-0)^2}=4$$ $$\sqrt{(at_1^2+a)^2}=4$$
$$\color{red}{|at_1^2+a|=4}\tag 1$$
similarly, the distance of the point $Q(at_2^2, 2at_2)$ from the focus $S(a, 0)$ is given as $$QS=\sqrt{(at_2^2-a)^2+(2at_2-0)^2}=9$$ $$\sqrt{(at_2^2+a)^2}=9$$
$$\color{blue}{|at_2^2+a|=9}\tag 2$$
then the equation of tangent at the point $P(at_1^2, 2at_1)$ is 
$$y-2at_1=\left[\frac{dy}{dx}\right]_{(at_1^2, 2at_1)}(x-at_1^2)$$
$$t_1y=x+at_1^2$$
similarly, the equation of the tangent at $Q(at_2^2, 2at_2)$ is 
$$t_2y=x+at_2^2$$
Now, solving the above two equations of the tangents, one should get intersection point $\left(at_1t_2, a(t_1+t_2)\right)$. 
Hence, the distance of the intersection point $\left(at_1t_2, a(t_1+t_2)\right)$ from the focus $(a, 0)$ is 
$$\sqrt{(at_1t_2-a)^2+(a(t_1+t_2)-0)^2}$$  $$=\sqrt{(at_1^2+a)(at_2^2+a)}$$ 
setting the values from (1) & (2), 
$$=\sqrt{(\color{red}{4})(\color{blue}{9})}=\color{red}{6}$$ 
