Question:
Let $L$ be a tangent line to the parabola $y^{2}=4 x-20$ at $(6,2)$. If $L$ is also a tangent to the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{b}=1$, then the value of $b$ is equal to :
Correct Option: , 2
Solution:
Tangent to parabola $2 y=2(x+6)-20$
$\Rightarrow y=x-4$
Condition of tangency for ellipse. $16=2(1)^{2}+b$
$\Rightarrow b=14$