JEE Main 2022 — Parabola Question with Solution

$$H:\frac{x^2}{4}-\frac{y^2}{4}=1$$

Focus $$(ae,0)$$

$$F\left(2\sqrt{2},0\right)$$

$$y=mx+c$$ passes through (1, 0)

$$0=m+C$$ ...... (i)

L is tangent to hyperbola

$$C=\pm \sqrt{4m^2-4}$$

$$-m=\pm \sqrt{4m^2-4}$$

$$m^2=4m^2-4$$

$$m=\frac{2}{\sqrt{3}}$$

$$C=\frac{-2}{\sqrt{3}}$$

$$T:y=\frac{2}{\sqrt{3}}x-\frac{2}{\sqrt{3}}$$

$$P:y^2=4x$$

$$y^2=4\left(\frac{\sqrt{3}y+2}{2}\right)$$

$$y^2-2\sqrt{3}y-4=0$$

Area

{1 \over 2}\left| {\matrix{ 0 & 0 \cr {{x_1}} & {{y_1}} \cr {2\sqrt 2 } & 0 \cr {{x_2}} & {{y_2}} \cr 0 & 0 \cr } } \right|

$$=\left|\frac{1}{2}\left(-2\sqrt{2}{y}_1+2\sqrt{2}{y}_2\right)\right|$$

$$=\sqrt{2}\left|y_2-y_1\right|=\sqrt{2}\sqrt{{(y_1+y_2)}^2-4y_1{y}_2}$$

$$=\sqrt{56}$$

$$=2\sqrt{14}$$