JEE Main 2021 — Ellipse Question with Solution

The equation of a tangent to a parabola $y^2=4a\left(x-h\right)$ at a point $\left(x_1, y_1\right)$ on it is $y{y}_1=4a\left(\left(\frac{x+x_1}{2}\right)-h\right).$

Hence, the equation of tangent to parabola $y^2=4x-20$ at $\left(6, 2\right)$ is $2y=2(x+6)-20$

$\Rightarrow y=x-4.$

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

Hence, by using the condition of tangency for ellipse$\frac{x^2}{2}+\frac{y^2}{b}=1$ and the line $y=x-4,$ we get

$16=2(1{)}^2+\mathrm{b}$

$\Rightarrow b=14.$