Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Let $P(3\, sec\, \theta ,2\, tan\, \theta )\; and \; Q(3\, sec\, \Phi ,2\, tan\, \Phi )\; where\, \theta +\Phi =\frac{\pi }{2},$ be two distinct points on the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ .Then the ordinate of the point of intersection of the normals at P and Q is :

Option 1)
$\frac{11}{3}$

Option 2)
$-\frac{11}{3}$

Option 3)
$\frac{13}{2}$

Option 4)
$-\frac{13}{2}$

Answers (1)

best_answer

As we learnt in

Parametric Coordinates -

$x=a\sec \Theta$

$y=b\tan \Theta$

- wherein

For the Hyperbola

$\frac{x^{2}}{a^{2}}- \frac {y^{2}}{b^{2}}= 1$ at Parametric $\Theta$

Parametric coordinate of Hyperbola is $\left ( a\sec \theta , b\tan \theta \right )$

Hence,

equation of normal is

$ax\cos \theta +by \cot \theta = a^{2}+b^{2}$

Normal at P and Q will be

$3x\cos \theta + 2y \cot \theta = 13$ .................................(1)

and $\:\: 3x \cos \phi + 2y \cot \phi = 13$ ......................... (2)

Solving for $y$

we get $\:\: y = \frac{-13}{2}$ .

Option 1)

$\frac{11}{3}$

This option is incorrect.

Option 2)

$-\frac{11}{3}$

This option is incorrect.

Option 3)

$\frac{13}{2}$

This option is incorrect.

Option 4)

$-\frac{13}{2}$

This option is correct.

Posted by

prateek

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 chemistry needs conceptual understanding; 5 tips to score high

anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon; paper pattern, exams from January 22

anam-khan

JEE Main Maths 2025: Five tips to score high marks in ‘toughest’ subject; most important chapters

Latest Question