JEE Coordinate (5)!

Geometry Level 4

Through the vertex of the parabola $y^2 = 4ax$ , two chords are drawn and the circle on these chords as diameters intersect at a point.

If $A$ and $B$ be the angles made with the $x$ -axis by tangents at the other ends of chords and $C$ be the angle made with the $x$ -axis by the line joining vertex of the parabola and point of intersection of circles, then $\cot(A)+\cot(B)+m\tan(C) = 0$ for some constant positive integer $m$ .

Find $m$ .

This is a part of my set Practice for JEE 2017!

3 2 8 5 7 6 1 4

1 solution

Tapas Mazumdar
Jan 9, 2018

The slope of the tangents calculated as:

$\begin{aligned} & y^2 = 4ax \\ \implies & 2y \dfrac{dy}{dx} = 4a \\ \implies & \dfrac{dy}{dx} = \dfrac{2a}{y} = \dfrac{2a}{2at_i} = \dfrac{1}{t_i} \ \ \ \ \ \ (i=1,2) \end{aligned}$

The equation of the two circles are:

$S_1 : x(x - at_1^2) + y(y - 2at_1) = 0 \\ S_2 : x(x - at_2^2) + y(y - 2at_2) = 0$

Equation of line joining the vertex of parabola to the intersection of the two circles is the common chord of the two circles given by,

$L : S_1 - S_2 = 0 \implies L : y = - \left( \dfrac{t_1 + t_2}{2} \right) x$

Using this we have,

$\begin{aligned} & \tan C = - \left( \dfrac{t_1 + t_2}{2} \right) \\ \implies & \tan C = - \left( \dfrac{(\tan A)^{-1} + (\tan B)^{-1}}{2} \right) \\ \implies & \tan C = - \left( \dfrac{\cot A + \cot B}{2} \right) \\ \implies & \boxed{\cot A + \cot B + 2 \tan C = 0} \end{aligned}$

JEE Coordinate (5)!

1 solution

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