1. ## hyperbola help!

$P(a\sec\theta, b\tan\theta)$ is a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. The ordinate at P and the tangent at P meet the asymptote at Q and R respectively. if the normal at P meets the x-axis at N prove that RQ and QN are perpendicular.

The ordinate at P is its y-value.
Asymptotes: $y=\pm\frac{b}{a}x$
when $y=b\tan\theta$, $x=\frac{a}{b}y=a\tan\theta$
$Q(a\tan\theta,b\tan\theta)$
equation of tangent $bx\sec\theta-ay\tan\theta=ab$
when $y=\frac{b}{a}x$
$bx(\sec\theta-\tan\theta)=ab$
$x=\frac{a}{\sec\theta-\tan\theta}$
$y=\frac{b}{a}(\frac{a}{\sec\theta-\tan\theta})=\frac{b}{\sec\theta-\tan\theta}$
$R(\frac{a}{\sec\theta-\tan\theta},\frac{b}{\sec\theta-\tan\theta})$
equation of normal $by\sec\theta+ax\tan\theta=(a^2+b^2)\tan\theta\sec\ theta$
when y=0, $x=\frac{a^2+b^2}{a}\sec\theta$
$N(\frac{a^2+b^2}{a}\sec\theta,0)$

gradient RQ $=\frac{b}{a}$
gradient QN $=\frac{b\tan\theta}{a\tan\theta-\frac{a^2+b^2}{a}\sec\theta}$
my problem is i can't get the product of the two gradients to be -1.
thanks!

2. You're good for $\theta = \frac{\pi}{2}$. Isn't that the point?

3. In the problem, it is stated that the ordinate at P and the tangent at P meet the asymptote at Q and R respectively.
I presume that the point Q lies on the asymptote y = bx/a, and R lies on the asymptote y = -bx/a.
The x co-ordinate of Q is the same as the x co-ordinate of P. Only ordinate will change. So the slope of RQ is not equal to b/a.

4. Originally Posted by sa-ri-ga-ma
In the problem, it is stated that the ordinate at P and the tangent at P meet the asymptote at Q and R respectively.
I presume that the point Q lies on the asymptote y = bx/a, and R lies on the asymptote y = -bx/a.
The x co-ordinate of Q is the same as the x co-ordinate of P. Only ordinate will change. So the slope of RQ is not equal to b/a.
If R intersects $y=-\frac{b}{a}x$, then $R(\frac{a}{\sec\theta+\tan\theta},-\frac{b}{\sec\theta+\tan\theta})$ which does not help.
$\frac{b(\sec^2\theta+\sec\theta)}{a(\sec^2\theta+\ sec\theta-2)}$