JEE Advanced 2022 — Paper 1 — Question 13

Let $P_{1}$ and $P_{2}$ be $\left(t_{1}^{2}, 2 t_{1}\right)$ and $\left(t_{2}^{2}, 2 t_{2}\right)$

$\Rightarrow \mathrm{P} \equiv\left(\mathrm{t}_{1} \mathrm{t}_{2}, \mathrm{t}_{1}+\mathrm{t}_{2}\right) \equiv(-2,1)$

$\Rightarrow t_{1}=2, t_{2}=-1$ or $t_{1}=-1, t_{2}=2$

$\Rightarrow P_{1}(4,4)$ and $P_{2}(1,-2)$ Slope of $\mathrm{SP}_{1}=\frac{4}{3}$

slope of $\mathrm{SP}_{2}=\infty$ Equation of $\mathrm{SP}_{1}$

$$y-0=\frac{4}{3}(x-1) \Rightarrow 4 x-3 y-4=0$$

Equation of $\mathrm{PQ}_{1}$ $y-1=-\frac{4}{3}(x+2) \Rightarrow 3 x+4 y+2=0$

$\therefore Q_{1}\left(\frac{2}{5},-\frac{4}{5}\right)$

Equation of $\mathrm{SP}_{2} \quad \mathrm{x}=1$

Equation of $\mathrm{PQ}_{2} \quad \mathrm{y}=1 \quad$

$\therefore \mathrm{Q}_{2}(1,1)$

$\therefore \mathrm{SQ}_{1}=1, \mathrm{Q}_{1} \mathrm{Q}_{2}=\frac{3 \sqrt{10}}{5}$

$\mathrm{PQ}_{1}=\sqrt{\left(-2-\frac{2}{5}\right)^{2}+\left(1+\frac{4}{5}\right)^{2}}=\sqrt{\frac{144}{25}+\frac{81}{25}}=3$ and $\mathrm{SQ}_{2}=\sqrt{(1-1)^{2}+(1-0)^{2}}=1$