Question

Let $a,b$and $\lambda$ be positive real numbers. Suppose $P$ is an end point of the latus rectum of the parabola $y^2=4\lambda$ , and suppose the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the point . If the tangents to the parabola and the ellipse at the point $P$ are perpendicular to each other, then the eccentricity of the ellipse is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{2}{5}$

Answer 1

The correct option is A

$\frac{1}{\sqrt{2}}$

Explanation for the correct option:

Finding the eccentricity of the ellipse.

Given: $a,b$and $\lambda$ be positive real numbers. $P$ is an endpoint of the latus rectum of the parabola $y^2=4\lambda$

tangent to parabola $y^2=4\lambda$ at $P(\lambda ,2\lambda )$

$2y\lambda =2\lambda (x+\lambda )$

and it's slope is $m_1=1$

tangent to ellipse at $P$

$\frac{\lambda x}{a^2}+\frac{2\lambda y}{b^2}=1$

it's slope is $m_2=-\left(\frac{b^2}{2a^2}\right)$

Since, Both tangents are perpendicular hence

$m_1{m}_2=-1$

$$\begin{gathered} \Rightarrow \frac{b^2}{2a^2}=1 \\ \Rightarrow \frac{a^2}{b^2}=\frac{1}{2} \end{gathered}$$

So, ecentricity is given by

$$\begin{gathered} e=\sqrt{1-\frac{a^2}{b^2}} \\ e=\sqrt{1-\frac{1}{2}} \\ \Rightarrow e=\frac{1}{\sqrt{2}} \end{gathered}$$

Hence the correct option is (A)