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  • There exist two points P and Q on the hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7

There exist two points P and Q on the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} such that \mathrm{O P \perp O Q, } where O is the origin, then the number of points in the xy-plane from where pair of perpendicular tangents can be drawn to the hyperbola is

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

Infinite


Answers (1)

best_answer

Let \mathrm{P\left(a \sec \theta_1, b \tan \theta_1\right)}  and \mathrm{Q\left(a \sec \theta_2, b \tan \theta_2\right)} be two points on the hyperbola

\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \, \, such\, \, that \, \, O P \perp O Q}

\mathrm{ \begin{aligned} & \Rightarrow \frac{b}{a} \frac{\tan \theta_1}{\sec \theta_1} \times \frac{b}{a} \frac{\tan \theta_2}{\sec \theta_2}=-1 \Rightarrow \sin \theta_1 \sin \theta_2=-\frac{a^2}{b^2} \\\\ & \Rightarrow a^2<b^2 . \end{aligned} }

The points in xy-plane from where perpendicular tangents are drawn to the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} lie on the director circle \mathrm{x^2+y^2=a^2-b^2.} For \mathrm{a^2<b^2}, the director circle does not exist. So, there is no point in xy-plane.

Posted by

Divya Prakash Singh

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