Question:
Let $L$ be a tangent line to the parabola $y^{2}=4 x-20$ at $(6,2)$. If $\mathrm{L}$ 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
$2 \mathrm{y}=2(\mathrm{x}+6)-20$
$\Rightarrow \mathrm{y}=\mathrm{x}-4$
Condition of tangency for ellipse.
$16=2(1)^{2}+b$
$\Rightarrow b=14$
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