JEE Main 2020 — Parabola Question with Solution

Equation of tangent to $y^2=4ax$ in slope Form is $y=mx+\frac{a}{m}$.

Comparing with given equation of tangent and parabola we get $\frac{1}{m}=4\Rightarrow m=\frac{1}{4}$.

Hence, line $y=\frac{x}{4}+4$ is tangent to $x^2=2by$ also.

Solving both, we get $x^2=2b\left(\frac{x}{4}+4\right)\Rightarrow 2x^2-bx-16b=0 ..\left(1\right)$.

Since the line is a tangent, so $\left(1\right)$ must have equal roots.

$\Rightarrow {\left(-b\right)}^2-4·2·\left(-16b\right)=0$

$\Rightarrow b^2+128b=0\Rightarrow b=0,-128$

But $b$ can't be zero.

So, $b=-128$