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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)


  • Option 2)


  • Option 3)


  • Option 4)



Answers (1)


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 )


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)


This option is incorrect.

Option 2)


This option is incorrect.

Option 3)


This option is incorrect.

Option 4)


This option is correct.

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