Question

If the normals at points of an ellipse, whose eccentric angles are $\alpha$, $\beta$, and $\gamma$, meet in a point, prove that
$\sin (\beta +\gamma )+\sin (\gamma +\alpha )+\sin (\alpha +\beta )=0$.
Hence, show that if $PQR$ be a maximum triangle inscribed in an ellipse, the normals at $P$, $Q$, and $R$ are concurrent.

Answer 1

The equation to the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let $A$ be the point $(a\cos \alpha ,b\sin \alpha )$ i.e., having eccentric angle $\alpha$;
Similarly $B$ be $(a\cos \beta ,b\sin \beta )$, C be $(a\cos \gamma ,b\sin \gamma )$ and D be $(a\cos \delta ,b\sin \delta )$; then the equation to AB is
$\frac{x}{a}\cos \frac{(\alpha +\beta )}{2}+\frac{y}{b}\sin \frac{(\alpha +\beta )}{2}=\cos \frac{\alpha -\beta }{2}....1$
and the equation to $CD$ is;-
$\frac{x}{a}\cos \frac{\gamma +\delta }{2}+\frac{y}{b}\sin \frac{\gamma +\delta }{2}=\cos \frac{\gamma -\delta }{2}$
Compare these lines with $l_{1}x+m_{1}y=1$ and $l_{2}x+m_{2}y=1$ and use the condition given by art $412$, i.e., $l_1{l}_2{a}^2=m_1{m}_2{b}^2=-1$
we get $\cos \frac{(\alpha +\beta )}{2}.\cos \frac{\gamma +\delta }{2}+\cos \frac{\alpha -\beta }{2}.\cos \frac{\gamma -\delta }{2}=0$
where,
$\cos \frac{\alpha +\beta +\gamma +\delta }{2}+\cos \frac{\gamma +\delta -\alpha -\beta }{2}+\cos \frac{\beta +\delta +\alpha -\gamma }{2}+\cos \frac{\delta +\alpha -\beta -\gamma }{2}=0\because \cos (-\theta )=\cos \theta$
$\alpha +\beta +\gamma +\delta =(2n+1)\overline{n}$
So, $\frac{\alpha +\beta +\gamma +\delta }{2}=(2n+1)\frac{\overline{n}}{2}$
So, $\frac{\alpha +\beta +\gamma +\delta }{2}=0$
Again, $\cos \left(\frac{\gamma +\delta -(\alpha +\beta )}{2}\right)=\cos (\frac{\overline{n}}{2}-(\alpha +\beta ))$
$=\sin (\alpha +\beta )$
Similarly other two terms reduced to,
$\sin (\beta +\gamma )$ and $\sin (\gamma +\alpha )$
So, the above equation becomes,
$\sin (\alpha +\beta )+\sin (\beta +\gamma )+\sin (\gamma +\alpha )=0$
Again the eccentric angles of $P,Q,R$ are
$\alpha ,\alpha +\frac{2\overline{n}}{3},\alpha +\frac{4\overline{n}}{3}$
As these angles satisfy the above relation, i.e, if we replace $\alpha ,\beta$ and $\gamma$ by these values respectively, the equation is satisfied.