Question

Chords at right angles are drawn through any point $P$ of the ellipse, and the line joining their extremities meets the normal in the point $Q$. Prove that $Q$ is the same for all such chords, its coordinates being $\frac{a^3{e}^2\cos \alpha }{a^2+b^2}$ and $\frac{-a^{2}b{e}^2\sin \alpha }{a^2+b^2}$.
Prove also that the major axis is the bisector of the angle $PCQ$, and that the locus of $Q$ for different positions of $P$ is the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}={\left(\frac{a^2-b^2}{a^2+b^2}\right)}^2$.

Answer 1

Let the equation to the normal at any point $(a\cos \alpha ,b\sin \alpha )$ of the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be
$ax\sec \alpha -by\csc \alpha =a^2-b^2$
Transferring the origin to the point$(a\cos \alpha ,b\sin \alpha )$
the equation to the ellipse becomes
$\frac{1}{a^2}{(x+a\cos \alpha )}^2+\frac{1}{b^2}{(y+b\sin \alpha )}^2=1$
$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{2x\cos \alpha }{a}+\frac{2y\cos \alpha }{b}=\mathrm{0....}\left(1\right)$
and the equation to the normal will be
$a(x+\cos \alpha )\sec \alpha -b(y+b\sin \alpha )\csc \alpha =a^2-b^2.....\left(2\right)$
Let PL and PM be the chords of the ellipse; then $\angle LPM=90°$and let the equation to the LM be $lx+my=\mathrm{1.....}\left(3\right)$
Making $\left(1\right)$ homogeneous by $\left(3\right)$ we have
$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{2x\cos \alpha }{a}(lx+my)+\frac{2y\sin \alpha }{b}(lx+my)=0$
$\therefore x^2(\frac{1}{a^2}+\frac{2l\cos \alpha }{a})+y^2(\frac{1}{b^2}+\frac{2m\sin \alpha }{b})+....=\mathrm{0.......}\left(4\right)$
Equation $\left(4\right)$ represents a pair of straight lines through the points of intersection of $\left(3\right)$ and $\left(1\right)$ and if, these lines are perpendicular then co efficient of $x^2+$ co efficient of $y^2=0$
i.e., $\frac{1}{a^2}+\frac{2l\cos \alpha }{a}+\frac{1}{b^2}+\frac{2m\sin \alpha }{b}=0$
$\frac{2}{ab}(bl\cos \alpha +am\sin \alpha )+\frac{a^2+b^2}{a^2{b}^2}=0$
or $bl\cos \alpha +am\sin \alpha =-\frac{a^2+b^2}{a^2{b}^2}.........\left(5\right)$
Again solving $\left(3\right)$ and $\left(1\right)$, the co ordinates of $Q(x,y)$ will be given by
$x=\frac{b\cos \alpha }{bl\cos \alpha +am\sin \alpha }=\frac{b\cos \alpha 2ba}{-(a^2+b^2)}$
$=-\frac{b^{2}a}{a^2+b^2}\cos \alpha$
and by $\left(2\right)$
$y=\frac{a}{b}x\tan \alpha$ or $y=-\frac{2a^{2}b}{a^2+b^2}\sin \alpha$
Changing to the original origin the co ordinates of Q are
$x=a\cos \alpha +\frac{-2a{b}^2\cos \alpha }{a^2+b^2}$
$=\frac{a(a^2-b^2)\cos \alpha }{a^2+b^2}=\frac{a^3{e}^2\cos \alpha }{a^2+b^2}$
$\frac{a^2-b^2}{a^2}=e^2$
and $y=b\sin \alpha +\frac{-2a^{2}b\sin \alpha }{a^2+b^2}$
$=-\frac{b(a^2-b^2)\sin \alpha }{a^2+b^2}=-\frac{a^{2}b{e}^2\sin \alpha }{a^2+b^2}$
which is constant because a,b,e and $\alpha$ are constants.
To find the locus of Q, we have to eliminate $\alpha$ by $\left(6\right)$ and $\left(7\right)$, we get
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{a^4{e}^4{\csc }^2\alpha +a^4{e}^4{\sin }^2\alpha }{{(a^2+b^2)}^2}$
$=\frac{a^4{e}^4}{{(a^2+b^2)}^2}={\left(\frac{a^2{e}^2}{a^2+b^2}\right)}^2$
and $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{{(a^2-b^2)}^2}{{(a^2+b^2)}^2}$ is the required locus.
Again the equation to CP is $y=\frac{b}{a}x\tan \alpha$ and that of CQ is $y=-\frac{b}{a}x\tan \alpha$
$\therefore CP$ and $CQ$ make equal angles with the axis and hene the major axis bisects the angle $PCQ$