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

$14$

Explanation for the correct option:

Finding the value of $b$:

Given equation of parabola is $y^2=4x–20$

We know the equation of the tangent to the parabola $y^2=4ax+c$ at $\left(x_{1,}{y}_1\right)$

$\Rightarrow y{y}_1=4a\left(\frac{x+x_1}{2}\right)+c$

Since $\left(x_{1,}{y}_1\right)$$=(6,2)$ is given, and here $a=1$

Therefore, the equation of tangent to this parabola

$$\begin{gathered} \Rightarrow y\times 2=2\left(x+6\right)-20 \\ \Rightarrow y=x-4 \\ \Rightarrow x-y-4=0....\left(i\right) \end{gathered}$$

Now we know the condition for which the line $y=mx+c....\left(ii\right)$ becomes tangent of ellipse $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1....\left(iii\right)$

The given equation of ellipse is $\frac{x^2}{2}+\frac{y^2}{b}=1...\left(iv\right)$

Now comparing equations we get $m=1\&c=-4$

And comparing equations we get $A^2=2\&B^2=b$

Applying $c^2=a^2{m}^2+b^2$ we get

$$\begin{gathered} 16=2+b \\ b=14 \end{gathered}$$

Hence, option (B) is the correct answer.