Question

If a straight line drawn through the focus S oi a. hyperbola, parallel to an asymptote, meet the curve in P, prove that SP is one quarter of the latus rectum.

Answer 1

Let the equation of the hyperbola be,
$\frac{l}{r}=1-e\cos \theta$
We know that if the asymptote to any hyperbola makes an angle of $\Phi ,$, $\tan \Phi =\frac{b}{a}$
Hence, $\cos \Phi =\frac{a}{\sqrt{a^2+b^2}}=\sqrt{\frac{a^2}{a^2+b^2}}$
$\cos \Phi =\sqrt{\frac{a^2}{a^2+a^2(e^2-1)}}=\sqrt{\frac{a^2}{a^2{e}^2}}$$=\frac{1}{e}....1$
Now let the line SP be drawn parallel to the asymptote. If it makes an angle $\alpha$ with the axis, then
$\alpha =\pi +\Phi$ or $\alpha -\pi =\Phi$
Hence, $\cos \Phi =\cos (\alpha -\pi )\Rightarrow \frac{1}{e}=-\cos \alpha$
or $\cos \alpha =-\frac{1}{e}$, since the point P lies on the conic $\frac{l}{r}=1-e\cos \theta$, we have
$r=SP=\frac{l}{1-e\cos \alpha }=\frac{1}{1+e.\frac{1}{e}}=\frac{l}{2}$
$=$ half of the semi-latus rectum
$=$one-quarter of the latus rectum