Test Index

Parabola

Section: Mathematics

© examsnet.com

Question : 9 of 31

Marks: +1, -0

Let

a, b

\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 theparabola and the ellipse at the point

P

are perpendicular to each other, then the eccentricity of the ellipse is

[JEE Adv 2020 P1]

$\frac{1}{\sqrt{2}}$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{2}{5}$

Solution:

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

=1

and equation of tangent at

P

to ellipse

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

is :

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

\therefore

Slope of tangent

= \frac{-b^{2}}{2 a^{2}}

\because 1 \cdot \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}}

∵ Ellipse

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

passes through

(\lambda, 2\lambda)

Hence

b^{2}>a^{2}

© examsnet.com

Go to Question:

Prev Question

Next Question