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QJEEM24Mae566: JEE Main 2024 Mathematics - Class 11 JEEM24Mae566 Solution: Let and be three points on the parabola and let the line…

Let A,BA, B and CC be three points on the parabola y2=6xy^2=6 x and let the line segment ABA B meet the line LL through CC parallel to the xx -axis at the point DD . Let MM and NN respectively be the feet of the perpendiculars from AA and BB on LL . Then (AM⋅BNCD)2\left(\frac{A M \cdot B N}{C D}\right)^2 is equal to __________.

Step-by-Step Verified Solution

Equation of ABA B

y(t1+t2)=2x+2at1t2y\left(t_1+t_2\right)=2 x+2 a t_1 t_2

JEE Main 2024 (Online) 9th April Evening Shift Mathematics - Parabola Question 19 English Explanation

amp; For D,y=2at3amp;⇒x=a(t1t3+t2t3−t1t2)amp;CD=∣a(t1t3+t2t3−t1t3)−at32∣amp;AM=∣2at1−2at3∣amp;BN=∣2at3−2at2∣amp;(AM⋅BNCD)2=(4a2(t1−t3)(t3−t2)a(t1t3+t2t3−t1t3−t32))2amp;=(4a2(t1−t3)(t3−t2)a(t1−t3)(t2−t3))2amp;=16a2=16⋅(32)2=36\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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