Question

Shew that the length of any focal chord of a conic is a third proportional to the transverse axis and the diameter parallel to the chord.

Answer 1

Let the conic be
$\frac{l}{r}=1-e\cos \theta$
If 'r' be the radius vector of any point, the diameter through the point (ssay x) will be $2r$. As we know,
$r^2=\frac{a^2{b}^2}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$
Hence, $x^2=4r^2=\frac{4a^2{b}^2}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }......1$
If $PS{P}^{{}^{\prime }}$ be nay chord clealr,
$PS=\frac{l}{1-e\cos \theta }$ and $P^{{}^{\prime }}S=\frac{l}{1-e\cos (\pi +\theta )}$
$P^{{}^{\prime }}S=\frac{l}{1+e\cos \theta }$
Let the length $PS{P}^{{}^{\prime }}$ be y; then
$y=\frac{l}{1-e\cos \theta }+\frac{l}{1+e\cos \theta }=\frac{2l}{1-e^2{\cos }^2\theta }....2$
Again in a conic, we have
$b^2=a^2(1-e^2)$
or $e^2=1-\frac{b^2}{a^2}=\frac{a^2-b^2}{a^2}$ and $2l=\frac{2b^2}{a}$
Putting in $2$, we get
$y=\frac{\frac{2b^2}{a}{a}^2}{a^2-(a^2-b^2){\cos }^2\theta }=\frac{2b^{2}a}{a^2(1-{\cos }^2\theta )+b^2{\cos }^2\theta }$
$\frac{2b^{2}a}{b^2{\cos }^2\theta +a^2{\cos }^2\theta }=\frac{1}{2a}\frac{4a^2{b}^2}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$
$=\frac{1}{2a}{x}^2$ by $1$
or $x^2=2ay$ or $\frac{2a}{x}=\frac{x}{y}$ or $2a:x::x:y$
i.e., the length of any focal chord of a conic is third proportional to the traverse axis and the diameter parallel to that chord.