JEE Main 2022 — Parabola Question with Solution

Equation of the tangent to the parabola can be written as $y=mx+\left(\frac{{\displaystyle 1}}{4m}\right)$ or $y-mx-\frac{1}{4m}=0$

Now, the perpendicular distance from $\left(0,0\right)$ to the tangent is equal to the radius of the given circle, so$OM=\sqrt{2}\Rightarrow \left|\frac{\frac{-1}{4m}}{\sqrt{1+m^2}}\right|=\sqrt{2}$

$\Rightarrow 2\left(16m^2\right)\left(1+m^2\right)=1 \Rightarrow 32m^4+32m^2-1=0$

$\Rightarrow m^2=\frac{-2^5+2^3\sqrt{16+2}}{2\times 2^5}$ ignoring the negative sign as it is square function.

Now $\left|8m_1{m}_2\right|=\left|8m^2\right|$

$=\left|\frac{-2^5+2^3\sqrt{16+2}}{2^3}\right|=3\sqrt{2}-4$