JEE Main 2022 — Parabola Question with Solution

Given parabola $$y^2=4x$$

$$\therefore$$ a = 1

Here, P, S, Q points are collinear.

$$\therefore$$ Slope of PS = Slope of QS

$$\Rightarrow \frac{2t_1-0}{t_1^2-1}=\frac{0-2t_2}{1-t_2^2}$$

$$\Rightarrow \frac{2t_1}{t_1^2-1}=\frac{2t_2}{t_2^2-1}$$

$$\Rightarrow t_1(t_2^2-1)=t_2(t_1^2-1)$$

$$\Rightarrow t_2^2{t}_1-t_1=t_1^2{t}_2-t_2$$

$$\Rightarrow t_2^2{t}_1-t_1^2{t}_2-t_1+t_2=0$$

$$\Rightarrow t_1{t}_2(t_2-t_1)+(t_2-t_1)=0$$

$$\Rightarrow (t_2-t_1)(t_1{t}_2+1)=0$$

As $$t_2-t_1\ne 0$$

$$\therefore$$ $$t_1{t}_2+1=0$$

$$t_1{t}_2=-1$$

Now, lenght of PQ

$$=\sqrt{{\left(t_1^2-t_2^2\right)}^2+{\left(2t_1-2t_2\right)}^2}$$

$$=\sqrt{{\left(t_1+t_2\right)}^2{\left(t_1-t_2\right)}^2+4{\left(t_1-t_2\right)}^2}$$

$$=\left(t_1-t_2\right)\sqrt{{\left(t_1+t_2\right)}^2+4}$$

$$=\left(t_1-t_2\right)\sqrt{t_1^2+t_2^2+2t_1{t}_2+4}$$

$$=\left(t_1-t_2\right)\sqrt{t_1^2+t_2^2+2(-1)+4}$$

$$=\left(t_1-t_2\right)\sqrt{t_1^2+t_2^2+2}$$

$$=\left(t_1-t_2\right)\sqrt{t_1^2+t_2^2-2(-1)}$$

$$=\left(t_1-t_2\right)\sqrt{t_1^2+t_2^2-2t_1{t}_2}$$

$$=\left(t_1-t_2\right)\sqrt{{\left(t_1-t_2\right)}^2}$$

$$={\left(t_1-t_2\right)}^2$$

Given, length of $$PQ={\left(t_1-t_2\right)}^2=6.25$$

$$\Rightarrow t_1-t_2=2.5$$

Now, Area of $$\Delta OPQ$$

= \left| {{1 \over 2}\left| {\matrix{ {t_1^2} & {2{t_1}} & 1 \cr {t_2^2} & {2{t_2}} & 1 \cr 0 & 0 & 1 \cr } } \right|} \right|

$$=\left|\frac{1}{2}\left(2t_2{t}_1^2-2t_1{t}_2^2\right)\right|$$

$$=\left|\frac{1}{2}\times 2t_1{t}_2\left(t_1-t_2\right)\right|$$

$$=\left|t_1{t}_2\times \left(t_1-t_2\right)\right|$$

$$=\left|-1\times 2.5\right|$$

$$=2.5$$

$$\therefore$$ 10$$\Delta$$OPQ =

$$=10\times \frac{25}{10}=25$$