A normal to the hyperbola, $4 x^2-9 y^2=36$ meets the coordinate axes $x$ and $y$ at $A$ and $B$, respectively. If the parallelogram $O A B P(O$ being the origin) is formed, then the locus of $P$ is

JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

A normal to the hyperbola,

4 x^2-9 y^2=36

meets the coordinate axes

x

and

y

A

and

B

, respectively. If the parallelogram

O A B P(O

being the origin) is formed, then the locus of

P

View answer and explanation

Correct option: C

9 x^2-4 y^2=169

Option C (

9 x^2-4 y^2=169

) is correct because it represents the locus of point P derived from the properties of the normal to the hyperbola and the parallelogram OABP.

Practice All Chapter MCQs →

a

y=x^3-a x

x^2 y=1-y

x y=1-y

C

y^3-3 x y+2=0

H

\mathrm{C}

x

V

C

y

y-e^{x y}+x=0

y = f\left( x \right)

x

x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)

\theta = \left( {2k + 1} \right)\pi ,\;k \in Z