$ \frac{x^2}{a^2}+ \frac{y^2}{b^2} =1$
is
$ \frac{xx'} {a^2} + \frac{yy'} {b^2} = \frac{x'^2}{a^2}+ \frac{y'^2}{b^2}$
where (x',y') are coordinates of midpoint
This is apparently true for all conics. Where is this coming from?
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(You meant to write ellipse?) In Joachimsthal's notation this is equivalent to $s_1=s_{11}$. – Jan-Magnus Økland Nov 26 '17 at 10:32
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From this book. For more on Joachimsthal's notation see this page.
Find the equation of the chord of $s=0$ whose midpoint is $P_1$.
Let $AA'$ be the chord and if possible let it meet the polar of $P_1$ in $P_2$. Then $AP_1=P_1A'$, $$AP_2=-P_2A'=A'P_2,$$ which is absurd. Therefore the chord cannot meet the polar of $P_1$. Hence it is parallel to the polar and has equation $s_1=k$. But it passes through $P_1$.
$\therefore$ $s_{11}=k$. Hence the chord is $$s_1=s_{11}.$$
Jan-Magnus Økland
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@Vrisk: Use that the polar are the points $P_2$ s.t. the cross-ratio $(P_1AP_2A')=-1$. – Jan-Magnus Økland Nov 26 '17 at 18:24
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1Could u please explain more on how do you prove that the polar of the midpoint is parallel to the chord, rest I understood ? – Sooraj S Nov 06 '19 at 14:43
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