Question

Let $a,b$ and $\lambda$ be positive real numbers. Suppose $P$ is an end point of the latus rectum of the parabola $y^2=4\lambda x$, and suppose the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the point $P$. 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

Answer: (A)

Equation of tangent to parabola $y^2=4\lambda x$ at $P(\lambda ,2\lambda )$ is :
$y\cdot 2\lambda =2\lambda (x+\lambda )$
$\therefore x-y+\lambda =0$
$\therefore$ Slope of tangent $=\frac{-b^2}{2a^2}$
$\because 1.\frac{-b^2}{2a^2}=-1$
$\Rightarrow \frac{b^2}{a^2}=2$
$\therefore$ Eccentricity of ellipse $=\sqrt{1-\frac{a^2}{b^2}}$
$=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$
$\because$ Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
passes through $(\lambda ,2\lambda )$
Hence $b^2>a^2$