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JEE Main 201325 AprilOnline

Exam: JEE Main
Year: 2013
Date: 25 April
Mode: Online
Subject: Mathematics
Chapter: Ellipse

JEE Main 2013 · 25 April · Online · Mathematics · Ellipse

A point on the ellipse, 4 x^2+9 y^2=36 , where the normal is...

A point on the ellipse,

4 x^2+9 y^2=36

, where the normal is parallel to the line,

4 x-2 y-5=0

, is :

Options

A $\left(\frac{9}{5}, \frac{8}{5}\right)$
B $\left(\frac{8}{5},-\frac{9}{5}\right)$
C $\left(-\frac{9}{5}, \frac{8}{5}\right)$
D $\left(\frac{8}{5}, \frac{9}{5}\right)$

Correct answer

C. $\left(-\frac{9}{5}, \frac{8}{5}\right)$

Step-by-step solution

Given ellipse is

4 x^2+9 y^2=36 ⇒ \frac{x^2}{9}+\frac{y^2}{4}=

1 Normal at the point is parallel to the line 4 x-2 y-5=0 Slope of normal

=2Slope of

tangent =

\frac{-1}{2} Point

of contact to ellip

se \frac{x^2}{a^2}

+

\frac{y^2}{b^2}=1

and

line is \left(\frac{a^2 m}{\sqrt{a^2 m^2+b^2}}, \frac{b}{\sqrt{a^2 m^2+b^2}}\right) Now,

a^2=9, b^2=4

\therefore \quad \text{ Point }

=\left(\frac{-9}{5},\frac{8}{5}\right)

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