Question

Let $L$ be a tangent line to the parabola $y^2=4x–20$ 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 :

• A. $20$
• B. $14$
• C. $16$
• D. $11$

Answer 1

The correct option is B $14$

Parabola Equation is $y^2=4x-20$
Tangent at $P(6,2)$ will be
$2y=4\left(\frac{x+6}{2}\right)-20$
$\Rightarrow 2y=2x+12-20$
$\Rightarrow x-y-4=0\cdots \left(1\right)$
This is also tangent to ellipse $\frac{x^2}{2}+\frac{y^2}{b}=1$
Applying $c^2=a^2{m}^2+b^2$
${(-4)}^2=\left(2\right)\left(1\right)+b$
$\Rightarrow b=14$