Mathematics · Parabola

JEE Main 2024 — 9 April, Shift 2 — Question 26

Let A,B\mathrm{A}, \mathrm{B} and C 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 x -axis at the point D . Let M and N 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 \qquad

Answer: 36

Numerical answer — enter this value.

Step-by-step solution

figure

mAB=mAD\mathrm{m}_{\mathrm{AB}}=\mathrm{m}_{\mathrm{AD}}

⇒2t1+t2=2a(t1−t3)at12−α\Rightarrow \quad \frac{2}{\mathrm{t}_{1}+\mathrm{t}_{2}}=\frac{2 \mathrm{a}\left(\mathrm{t}_{1}-\mathrm{t}_{3}\right)}{\mathrm{at}_{1}^{2}-\alpha}

⇒at12−α=a{t12−t1t3+t1t2−t2t3}\Rightarrow \quad \mathrm{at}_{1}^{2}-\alpha=\mathrm{a}\left\{\mathrm{t}_{1}^{2}-\mathrm{t}_{1} \mathrm{t}_{3}+\mathrm{t}_{1} \mathrm{t}_{2}-\mathrm{t}_{2} \mathrm{t}_{3}\right\}

⇒α=a(t1t3+t2t3−t1t2)\Rightarrow \quad \alpha=\mathrm{a}\left(\mathrm{t}_{1} \mathrm{t}_{3}+\mathrm{t}_{2} \mathrm{t}_{3}-\mathrm{t}_{1} \mathrm{t}_{2}\right)

AM=∣2a(t1−t3)∣,BN=∣2a(t2−t3)∣A M=\left|2 \mathrm{a}\left(\mathrm{t}_{1}-\mathrm{t}_{3}\right)\right|, \quad \mathrm{BN}=\left|2 \mathrm{a}\left(\mathrm{t}_{2}-\mathrm{t}_{3}\right)\right|,

CD=∣at32−α∣\mathrm{CD}=\left|\mathrm{at}_{3}^{2}-\alpha\right|

CD=∣at32−a(t1t3+t2t3−t1t2)∣\mathrm{CD}=\left|\mathrm{at}_{3}^{2}-\mathrm{a}\left(\mathrm{t}_{1} \mathrm{t}_{3}+\mathrm{t}_{2} \mathrm{t}_{3}-\mathrm{t}_{1} \mathrm{t}_{2}\right)\right|

=a∣t32−t1t3−t2t3+t1t2∣\quad=\mathrm{a}\left|\mathrm{t}_{3}^{2}-\mathrm{t}_{1} \mathrm{t}_{3}-\mathrm{t}_{2} \mathrm{t}_{3}+\mathrm{t}_{1} \mathrm{t}_{2}\right|

=a∣t3(t3−t1)−t2(t3−t1)∣\quad=\mathrm{a}\left|\mathrm{t}_{3}\left(\mathrm{t}_{3}-\mathrm{t}_{1}\right)-\mathrm{t}_{2}\left(\mathrm{t}_{3}-\mathrm{t}_{1}\right)\right|

CD=a∣(t3−t2)(t3−t1)∣\mathrm{CD}=\mathrm{a}\left|\left(\mathrm{t}_{3}-\mathrm{t}_{2}\right)\left(\mathrm{t}_{3}-\mathrm{t}_{1}\right)\right|

(AM⋅BNCD)2={2a(t1−t3)⋅2a(t2−t3)a(t3−t2)(t3−t1)}2\left(\frac{\mathrm{AM} \cdot \mathrm{BN}}{\mathrm{CD}}\right)^{2}=\left\{\frac{2 \mathrm{a}\left(\mathrm{t}_{1}-\mathrm{t}_{3}\right) \cdot 2 \mathrm{a}\left(\mathrm{t}_{2}-\mathrm{t}_{3}\right)}{\mathrm{a}\left(\mathrm{t}_{3}-\mathrm{t}_{2}\right)\left(\mathrm{t}_{3}-\mathrm{t}_{1}\right)}\right\}^{2}

16a2=16×94=3616 \mathrm{a}^{2}=16 \times \frac{9}{4}=36

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Parabola
Topic
Chords connected with a Parabola