Question

If a chord of a conic section subtend a constant angle at the focus, prove that the locus of the point where it meets the internal bisector of the angle 2a is the conic section
$\frac{l\cos \delta }{r}=1-e\cos \delta \cos \theta .$

Answer 1

Let the conic be $\frac{l}{r}=1-\cos \theta$ having S as its focus. If AB be any chord of it such that $\angle ASB=2\alpha$, SP be the bisector of the $\angle ASB$, meeting AB in P, and the vectorial angle of P be $\beta$, clearly the vectorial angles of A and B will be $(\beta +\alpha )$ and $(\beta -\alpha )$
Hence the equation of AB will be
$\frac{l}{r}=\sec \alpha \cos (\theta -\beta )-e\cos \theta ....1$
Solving $1$ with the vectorial angle of P i.e., $\beta$, we get the locus of P as
$\frac{l}{r}=\sec \alpha \cos (\beta -\beta )-e\cos \theta$
or $\frac{l}{r}=\frac{1}{\cos \alpha }-e\cos \theta (\because \cos 0=1)$
or $\frac{l\cos \alpha }{r}=1-e\cos \theta \cos \alpha$