Question

If $P(asec\alpha ,btan\alpha )$ and $Q(asec\beta ,btan\beta )$ are two points on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $\alpha -\beta =2\theta$ (a costant), then PQ touches the hyperbola

• A. $\frac{x^2}{a^{2}se{c}^2\theta }-\frac{y^2}{b^2}=1$
• B. $\frac{x^2}{a^2}-\frac{y^2}{b^{2}se{c}^2\theta }=1$
• C. $\frac{x^2}{a^2}-\frac{y^2}{b^2}=co{s}^2\theta$
• D. none of these

Answer 1

The correct option is A $\frac{x^2}{a^{2}se{c}^2\theta }-\frac{y^2}{b^2}=1$

The equation of chord $PQ$ is $\frac{x}{a}\cos \left(\frac{\alpha -\beta }{2}\right)-\frac{y}{b}\sin \left(\frac{\alpha +\beta }{2}\right)=\cos \left(\frac{\alpha +\beta }{2}\right)$
$\Rightarrow \frac{x}{a}\cos \left(\frac{\alpha -\beta }{2}\right)-\frac{y}{b}\sin \left(\frac{\alpha +\beta }{2}\right)=\cos \left(\frac{\alpha +\beta }{2}\right)$
$\Rightarrow \frac{x^{}}{a^{}\sec \theta }\sec \left(\frac{\alpha +\beta }{2}\right)-\frac{y}{b}\tan \left(\frac{\alpha +\beta }{2}\right)=1$
Clearly, it touches the hyperbola
$\frac{x^2}{a^2{\sec }^2\theta }-\frac{y^2}{b^2}=1$