JEE MAIN - Mathematics (2024 - 9th April Evening Shift - No. 22)
Let $$A, B$$ and $$C$$ be three points on the parabola $$y^2=6 x$$ and let the line segment $$A B$$ meet the line $$L$$ through $$C$$ parallel to the $$x$$-axis at the point $$D$$. Let $$M$$ and $$N$$ respectively be the feet of the perpendiculars from $$A$$ and $$B$$ on $$L$$. Then $$\left(\frac{A M \cdot B N}{C D}\right)^2$$ is equal to __________.
Answer
36
Explanation
Equation of $$A B$$
$$y\left(t_1+t_2\right)=2 x+2 a t_1 t_2$$
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$$\begin{aligned} & \text { For } D, y=2 a t_3 \\ & \Rightarrow x=a\left(t_1 t_3+t_2 t_3-t_1 t_2\right) \\ & C D=\left|a\left(t_1 t_3+t_2 t_3-t_1 t_3\right)-a t_3^2\right| \\ & A M=\left|2 a t_1-2 a t_3\right| \\ & B N=\left|2 a t_3-2 a t_2\right| \\ & \left(\frac{A M \cdot B N}{C D}\right)^2=\left(\frac{4 a^2\left(t_1-t_3\right)\left(t_3-t_2\right)}{a\left(t_1 t_3+t_2 t_3-t_1 t_3-t_3^2\right)}\right)^2 \\ & =\left(\frac{4 a^2\left(t_1-t_3\right)\left(t_3-t_2\right)}{a\left(t_1-t_3\right)\left(t_2-t_3\right)}\right)^2 \\ & =16 a^2=16 \cdot\left(\frac{3}{2}\right)^2=36 \end{aligned}$$
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