JEE Main 2023 — Parabola Question with Solution

$$P\equiv \left(\frac{A}{m^2},\frac{2A}{m}\right)$$ where $$\left(A=\frac{3}{4},m=\frac{-1}{2}\right)$$

& $$Q,R=\left(\mp \frac{a^2{m}_1}{a^2{m}_1^2+b^2},\frac{\mp .b^2}{\sqrt{a^2{m}_1^2+b^2}}\right)$$

Where $$a^2=4,b^2=1$$ and $$m_1=1$$

$$\therefore$$ $$P\equiv (3,-3)$$

$$Q\equiv \left(\frac{-4}{\sqrt{5}},\frac{-1}{\sqrt{5}}\right)$$ & $$R\left(\frac{4}{\sqrt{5}},\frac{1}{\sqrt{5}}\right)$$

Area = {1 \over 2}\left| {\matrix{ 3 & { - 3} & 1 \cr {{{ - 4} \over {\sqrt 5 }}} & {{{ - 1} \over {\sqrt 5 }}} & 1 \cr {{4 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} & 1 \cr } } \right| = {1 \over {10}}\left| {\matrix{ 3 & { - 3} & 1 \cr { - 4} & { - 1} & {\sqrt 5 } \cr 0 & 0 & {2\sqrt 5 } \cr } } \right|

$$=\frac{2\sqrt{5}}{10}(-15)=3\sqrt{5}$$