Question

Through the vertex 𝑂 of the parabola, 𝑦2 = 4𝑎𝑥 two chords 𝑂𝑃 & 𝑂𝑄 are drawn and the circles on 𝑂𝑃 & 𝑂𝑄 as diameters intersect in 𝑅. If 1,2 &  are the angles made with the axis by the tangents at 𝑃 & 𝑄 on the parabola & by 𝑂𝑅 then the value of, cot1 + cot2 =

Answer 1

-2tanϕ

Answer 2

To solve the problem, we follow these steps:

1. Parametric Coordinates:
   Let the vertex of the parabola $y^2 = 4ax$ be $O(0,0)$.
   Let the parametric coordinates of points $P$ and $Q$ on the parabola be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$, respectively.

2. Equations of Circles:
   The equation of a circle with $OP$ as diameter (endpoints $O(0,0)$ and $P(at_1^2, 2at_1)$) is given by:
   $(x - 0)(x - at_1^2) + (y - 0)(y - 2at_1) = 0$
   $C_1: x^2 + y^2 - at_1^2 x - 2at_1 y = 0 \quad (1)$

   Similarly, the equation of a circle with $OQ$ as diameter (endpoints $O(0,0)$ and $Q(at_2^2, 2at_2)$) is:
   $(x - 0)(x - at_2^2) + (y - 0)(y - 2at_2) = 0$
   $C_2: x^2 + y^2 - at_2^2 x - 2at_2 y = 0 \quad (2)$

3. Equation of Common Chord OR:
   The circles intersect at $O(0,0)$ and $R$. The equation of the common chord $OR$ is obtained by subtracting equation (2) from equation (1):
   $(x^2 + y^2 - at_1^2 x - 2at_1 y) - (x^2 + y^2 - at_2^2 x - 2at_2 y) = 0$
   $(-at_1^2 + at_2^2)x + (-2at_1 + 2at_2)y = 0$
   $a(t_2^2 - t_1^2)x + 2a(t_2 - t_1)y = 0$
   Since $P$ and $Q$ are distinct points, $t_1 \neq t_2$. We can divide by $a(t_2 - t_1)$:
   $(t_2 + t_1)x + 2y = 0$
   $2y = -(t_1 + t_2)x$
   $y = -\frac{t_1 + t_2}{2}x$

4. Angle $\phi$ of OR:
   The slope of the line $OR$ is $m_{OR} = \tan\phi$.
   From the equation of $OR$, $m_{OR} = -\frac{t_1 + t_2}{2}$.
   So, $\tan\phi = -\frac{t_1 + t_2}{2}$.
   This implies $t_1 + t_2 = -2\tan\phi$.

5. Angles $\theta_1$ and $\theta_2$ of Tangents:
   The equation of the tangent to the parabola $y^2 = 4ax$ at a point $(at^2, 2at)$ is $ty = x + at^2$.
   For point $P(at_1^2, 2at_1)$, the tangent is $t_1 y = x + at_1^2$.
   The slope of the tangent at $P$ is $m_P = \frac{1}{t_1}$.
   Given that this tangent makes an angle $\theta_1$ with the axis, $m_P = \tan\theta_1$.
   So, $\tan\theta_1 = \frac{1}{t_1}$, which means $\cot\theta_1 = t_1$.

   Similarly, for point $Q(at_2^2, 2at_2)$, the tangent is $t_2 y = x + at_2^2$.
   The slope of the tangent at $Q$ is $m_Q = \frac{1}{t_2}$.
   Given that this tangent makes an angle $\theta_2$ with the axis, $m_Q = \tan\theta_2$.
   So, $\tan\theta_2 = \frac{1}{t_2}$, which means $\cot\theta_2 = t_2$.

6. Calculate $\cot\theta_1 + \cot\theta_2$:
   We need to find the value of $\cot\theta_1 + \cot\theta_2$.
   Substituting the expressions from step 5:
   $\cot\theta_1 + \cot\theta_2 = t_1 + t_2$
   From step 4, we know that $t_1 + t_2 = -2\tan\phi$.
   Therefore, $\cot\theta_1 + \cot\theta_2 = -2\tan\phi$.

The final answer is $\boxed{-2\tan\phi}$.