$\displaystyle P(a\sec\theta, b\tan\theta)$ is a point on the hyperbola $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. 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: $\displaystyle y=\pm\frac{b}{a}x$

when $\displaystyle y=b\tan\theta$, $\displaystyle x=\frac{a}{b}y=a\tan\theta$

$\displaystyle Q(a\tan\theta,b\tan\theta)$

equation of tangent $\displaystyle bx\sec\theta-ay\tan\theta=ab$

when $\displaystyle y=\frac{b}{a}x$

$\displaystyle bx(\sec\theta-\tan\theta)=ab$

$\displaystyle x=\frac{a}{\sec\theta-\tan\theta}$

$\displaystyle y=\frac{b}{a}(\frac{a}{\sec\theta-\tan\theta})=\frac{b}{\sec\theta-\tan\theta}$

$\displaystyle R(\frac{a}{\sec\theta-\tan\theta},\frac{b}{\sec\theta-\tan\theta})$

equation of normal $\displaystyle by\sec\theta+ax\tan\theta=(a^2+b^2)\tan\theta\sec\ theta$

when y=0, $\displaystyle x=\frac{a^2+b^2}{a}\sec\theta$

$\displaystyle N(\frac{a^2+b^2}{a}\sec\theta,0)$

gradient RQ $\displaystyle =\frac{b}{a}$

gradient QN $\displaystyle =\frac{b\tan\theta}{a\tan\theta-\frac{a^2+b^2}{a}\sec\theta}$

my problem is i can't get the product of the two gradients to be -1.

thanks!