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 $=\frac{-b^{2}}{2 a^{2}}$
$\because 1 .\frac{-b^{2}}{2 a^{2}}=-1$
$\Rightarrow \frac{b^{2}}{a^{2}}=2$
$\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 $b^{2}>a^{2}$