Question

Two equal ellipses, of eccentricity e, are placed with their axes at right angles and they have one focus S in common ; if PQ be a common tangent, show that the angle PSQ is equal to $2{\sin }^{-1}\frac{e}{\sqrt{2}}.$

Answer 1

Let $\frac{l}{r}=1-e\cos \theta$ be the equation of one ellipse.
Then the equations of the ellipse having a focus common, same eccentricity and axis at right angles to first will be
$\frac{l}{r}=1-e\sin \theta$
The equation of tangent at point ${}^{\prime }{\alpha }^{\prime }$ on the first ellipse will be
$\frac{l}{r}=\cos (\theta -\alpha )-e\cos \theta$
$\frac{l}{r}=\cos \theta (\cos \alpha -e)+\sin \theta \sin \alpha .....1$
Similarly the equation of the tangent at some point $\beta$ of the second ellipse will be
$\frac{l}{r}=\cos (\theta -\beta )-e\sin \theta$
or $\frac{l}{r}=\cos \theta \cos \beta +\sin \theta (\sin \beta -e)....2$
Since the lines given by $1$ and $2$ are the same comparing we get
$\cos \alpha -e=\cos \beta$ ...... $\left(3\right)$ and $\sin \beta -e=\sin \alpha$ ..... $\left(4\right)$
Subtracting we get $\cos \alpha -\cos \beta =\sin \beta -\sin \alpha$
or $2\sin \frac{\alpha +\beta }{2}\sin \frac{\beta -\alpha }{2}=2\cos \frac{\alpha +\beta }{2}\sin \frac{\beta -\alpha }{2}$
or $\tan \frac{\alpha +\beta }{2}=1$; hence $\frac{\alpha +\beta }{2}=\frac{\pi }{4}$
Adding $3$ and $4$, we have
$\cos \alpha -\cos \beta -\sin \alpha +\sin \beta =2e$
Dividing the whole equation by $\sqrt{2}$ we get;-
$\sin (\frac{\pi }{4}-\alpha )+\sin (\beta -\frac{\pi }{4})=\frac{2e}{\sqrt{2}}=\sqrt{2}e$
or $2\sin \frac{\beta -\alpha }{2}\cos (\frac{\alpha +\beta }{2}-\frac{\pi }{4})=\sqrt{2}e$
or $\sin \frac{\beta -\alpha }{2}=\frac{\sqrt{2}e}{2}=\frac{1}{\sqrt{2}}e(\because \frac{\alpha +\beta }{2}=\frac{\pi }{4})$
$\beta -\alpha =2{\sin }^{-1}\left(\frac{1}{\sqrt{2}}e\right)$