Mathematics · Parabola

JEE Main 2024 — 6 April, Shift 1 — Question 25

Let $\mathrm{L}_{1}, \mathrm{~L}_{2}$ be the lines passing through the point $\mathrm{P}(0,1)$ and touching the parabola $9 x^{2}+12 x+18 y-14=0$ .

Let $Q$ and $R$ be the points on the lines $L_{1}$ and $L_{2}$ such that the $\triangle P Q R$ is an isosceles triangle with base QR .

If the slopes of the lines $Q R$ are $m_{1}$ and $m_{2}$ . then $16\left(m_{1}^{2}+m_{2}^{2}\right)$ is equal to $\qquad$

Answer: 68

Numerical answer — enter this value.

$\quad 9 \mathrm{x}^{2}+12 \mathrm{x}+4=-18(\mathrm{y}-1)$

$(3 x+2)^{2}=-18(y-1)$

$\left(x+\frac{2}{3}\right)^{2}=-2(y-1)$

$y=m x+1$

$\left(x+\frac{2}{3}\right)^{2}=-2(y-1)$

$(3 x+2)^{2}=-18 m x$

$9 \mathrm{x}^{2}+(12+18 \mathrm{~m}) \mathrm{x}+4=0$

$4(6+9 \mathrm{~m})^{2}=4(36)$

$6+9 m=6,-6$

$\mathrm{m}=0, \frac{-4}{3}$

$\tan \theta=-\frac{4}{3}$

$\frac{2 \tan \frac{\theta}{2}}{1-\tan ^{2} \frac{\theta}{2}}=\frac{-4}{3}$

$\left(\tan \frac{\theta}{2}-2\right)\left(2 \tan \frac{\theta}{2}+1\right)=0$

$\tan \frac{\theta}{2}=2, \frac{-1}{2}$

$\mathrm{m}_{\mathrm{QR}}=\tan \left(90+\frac{\theta}{2}\right)$

$=-\cot \frac{\theta}{2}$

$\mathrm{m}_{1}=\frac{-1}{2} ,\quad \mathrm{~m}_{2}=\frac{-1}{-1 / 2}=2$

$16\left(\mathrm{~m}_{1}^{2}+\mathrm{m}_{2}^{2}\right)=16\left(\frac{1}{4}+4\right)$ $=4+64=68$

\boxed{68}

Answer key and solution verified before publishing.

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