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Let \mathrm{P(a \sec \theta, b \tan \theta)} and \mathrm{Q(a \sec \phi, b \tan \phi)}, where \mathrm{\theta+\phi=\pi / 2,} be two points on the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}. If (h, k) is the point of intersection of the normals at P and Q, then k is equal to

Option: 1

\mathrm{\frac{a^2+b^2}{a}}


Option: 2

\mathrm{-\left(\frac{a^2+b^2}{a}\right)}


Option: 3

\mathrm{\frac{a^2+b^2}{b}}


Option: 4

\mathrm{-\left(\frac{a^2+b^2}{b}\right)}


Answers (1)

best_answer

Equation of the normal to the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}

at the point \mathrm{(a \sec \alpha, b \tan \alpha)} is given by \mathrm{a x \cos \alpha+b y \cot \alpha=a^2+b^2}

Normal at P and Q are respectively \mathrm{a x \cos \theta+b y \cot \theta=a^2+b^2}

and \mathrm{a x \cos \phi+b y \cot \phi=a^2+b^2}

where \mathrm{\phi=\frac{\pi}{2}-\theta}and these pass through (h, k).

\mathrm{\therefore a h \cos \theta+b k \cot \theta=a^2+b^2\, \, and \, \, a h \sin \theta+b k \tan \theta=a^2+b^2}

Eliminating h, we get

\mathrm{ \begin{aligned} & b k(\cot \theta \sin \theta-\tan \theta \cos \theta)=\left(a^2+b^2\right)(\sin \theta-\cos \theta) \\ \Rightarrow \quad & k=-\left(a^2+b^2\right) / b \end{aligned} }

Posted by

himanshu.meshram

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