JEE Main 2021 — Parabola Question with Solution

Equation of the parabola is ${\left(x-\frac{1}{2}\right)}^2=\left(y-\frac{3}{4}\right) ...\left(1\right)$

Put $x=\frac{-1}{2}$ in equation $\left(1\right)$, we get $y=\frac{7}{4}$

Hence, point $P$ is $P\left(\frac{-1}{2},\frac{7}{4}\right)$

Now, differentiate equation $\left(1\right)$ on both sides, we get $y^{\text{'}}=2\left(x-\frac{1}{2}\right)$

At point $P\left(\frac{-1}{2},\frac{7}{4}\right)$, the slope of tangent is $-2$.

Therefore, the slope of normal is $\frac{1}{2}$

Equation of normal at $P\left(\frac{-1}{2}, \frac{7}{4}\right)$ is $x=2 y-4 ...\left(2\right)$

On solving equations $\left(1\right) \& \left(2\right)$, we get the intersection point $Q\left(2, 3\right)$

$\therefore (\mathrm{PQ}{)}^2={\left(2+\frac{1}{2}\right)}^2+{\left(3-\frac{7}{4}\right)}^2$

$=\frac{25}{4}+\frac{25}{16}=\frac{125}{16}$