JEE Main 2021 — Hyperbola Question with Solution

A line $y=mx+c$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ if $c^2=a^2{m}^2-b^2.$

Given that, tangent to hyperbola $\frac{x^2}{3}-\frac{y^2}{3}=1$ is $2x+y=k$ or $y=-2x+k$

Thus, we have, slope $m=-2, c= k \& a^2=b^2=3$

$k^2=3{\left(-2\right)}^2-3$

$\Rightarrow k^2=9$

Given $k>0, \Rightarrow k=3.$

Thus, the equation of the tangent to the hyperbola is $y=-2x+3.$

Given this line is also a tangent to the parabola, $y^2=\alpha x$ and a line $y=mx+c$ is tangent to the parabola $y^2=4Ax,$ if $c=\frac{A}{m}$

Thus, we have $3=\frac{\left({\displaystyle \frac{\alpha }{4}}\right)}{-2}$

$\Rightarrow \frac{\alpha }{-8}=3$

$\Rightarrow \alpha =-24.$