Question

A straight line is drawn parallel to the conjugate axis of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to meet it and the conjugate hyperbola respectively in the point $P$ and $Q$. The normals at $p$ and $Q$ to the curves meet on

• A. $x-axis$
• B. $y-axis$
• C. $y=x$
• D. $y=-x$

Answer 1

The correct option is A $x-axis$

Equation of hyperbola $\to \frac{x^2}{9^2}-\frac{y^2}{6^2}=1$

Conjugate hyperbola $\to \frac{y^2}{6^2}-\frac{x^2}{9^2}=1$

Line $||$ to conjugate axis of hyperbola $(y-axis)$ is drawn to meet conjugate axis at $P$ & $Q$ which are symmetric points about $x-$axis then

$P(a\tan \theta ,b\sin theta)$

$Q(a\tan \theta ,-b\sin \theta )$

Normal at $P\to \frac{ax}{\tan \theta }+\frac{6y}{\sec \theta }=a^2+b^2$

Normal at $Q\to \frac{ax}{\tan \theta }-\frac{6y}{\sec \theta }=a^2+b^2$

Let us find out interrection

$\frac{b}{\sec \theta }=\frac{-6y}{\sec \theta }$

$y=0$

$9+$ lies on $x-$axis

$A$ is correct