Circling a parabola!

Let $P \equiv (a{t_1}^2 , 2at_1)$ and $Q \equiv (a{t_2}^2 , 2at_2)$ be the two parametric points on the parabola $y^2 = 4ax$ .

Now,

Equation of circle ( $OP$ as diameter) $C_1 \equiv x(x-a{t_1}^2) + y(y-2at_1) = 0 \implies x^2 + y^2 - a{t_1}^2 x - 2at_1 y = 0$

Equation of circle ( $OQ$ as diameter) $C_2 \equiv x(x-a{t_2}^2) + y(y-2at_1) = 0 \implies x^2 + y^2 - a{t_2}^2 x - 2at_2 y = 0$

Since $OR$ is a common chord for $C_1$ and $C_2$ , therefore its equation is:

$C_1 - C_2 \equiv a{t_2}^2 x - a{t_1}^2 x + 2at_2 y - 2at_1 y = 0 \implies (t_1 + t_2) x + 2y = 0$

whose gradient is given by,

$\tan \varphi = - \dfrac{(t_1 + t_2)}{2} \qquad \cdots(1)$

Now,

Gradient of tangent to the parabola $y^2 = 4ax$ at $t_1 = \dfrac{1}{t_1}$

Gradient of tangent to the parabola $y^2 = 4ax$ at $t_2 = \dfrac{1}{t_2}$

So,

$\tan \theta_1 = \dfrac{1}{t_1} \implies \cot \theta_1 = t_1 \\ \tan \theta_2 = \dfrac{1}{t_2} \implies \cot \theta_2 = t_2$

Thus, from $(1)$ ,

$\tan \varphi = - \dfrac{(\cot \theta_1 + \cot \theta_2)}{2} \\ \therefore \cot \theta_1 + \cot \theta_2 = \boxed{-2 \tan \varphi}$

Did it by generalisation