Question

Show 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

Hint: First of all, consider the general equation of the conic and consider a chord on the conic and find out its length. Convert the obtained fractions into ratio and proportions to prove the problem.

Complete step-by-step answer:
Let the conic be
$\dfrac{l}{r} = 1 - e\cos \theta$

Where $\angle PAS = \theta$
Here $r$ is the radius vector of any point and the diameter through the point (say $P$) will be $2r$.
As we know that ${r^2} = \dfrac{{{a^2}{b^2}}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\varphi }}$ where $a$ is the length of the major axis and $b$ is the length of the minor axis.
Hence
$\Rightarrow {x^2} = {\left( {2r} \right)^2} = 4{r^2} = \dfrac{{4{a^2}{b^2}}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}.........................................\left( 1 \right)$
Let $PSP'$ be any chord on the conic then we have
$PS = \dfrac{l}{{1 - e\cos \theta }}$ and $P'S = \dfrac{l}{{1 - e\cos \left( {\pi + \theta } \right)}} = \dfrac{l}{{1 + e\cos \theta }}$
Let the length of the chord $PSP'$ be $y$, then we have

$$\Rightarrow y = PS + P'S \\\ \Rightarrow y = \dfrac{l}{{1 - e\cos \theta }} + \dfrac{l}{{1 + e\cos \theta }} = \dfrac{{2l}}{{1 - {e^2}{{\cos }^2}\theta }}..............................\left( 2 \right) \\\$$

Again, in a conic we have
${b^2} = {a^2}\left( {1 - {e^2}} \right)$
Which can be written as
$\Rightarrow {e^2} = 1 - \dfrac{{{b^2}}}{{{a^2}}} = \dfrac{{{a^2} - {b^2}}}{{{a^2}}}{\text{ and }}2l = \dfrac{{2{b^2}}}{a}.............................\left( 3 \right)$
Substituting (3) in (2), we get

$$\Rightarrow y = \dfrac{{\dfrac{{2{b^2}}}{a}{a^2}}}{{{a^2} - \left( {{a^2} - {b^2}} \right){{\cos }^2}\theta }} \\\ \Rightarrow y = \dfrac{{2{b^2}a}}{{{a^2}\left( {1 - {{\cos }^2}\theta } \right) + {b^2}{{\cos }^2}\theta }} \\\ \Rightarrow y = \dfrac{{2{b^2}a}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\cos }^2}\theta }} \\\ \Rightarrow y = \dfrac{1}{{2a}} \times \dfrac{{4{a^2}{b^2}}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }} \\\$$

From equation (1), we have

$$\Rightarrow y = \dfrac{1}{{2a}}{x^2} \\\ \therefore {x^2} = 2ay{\text{ or }}\dfrac{{2a}}{x} = \dfrac{x}{y} \\\$$

By converting these into ratio and proportion, we get
$2a:x$ is in proportion to $x:y$
i.e., the length of any focal chord of a conic is third proportional to the transverse axis and the diameter parallel to that chord.
Hence proved.

Note: The general equation of a conic is given by $\dfrac{l}{r} = 1 - e\cos \theta$ where $e$ is the eccentricity of the conic. Here the point $S$ is the focus of the conic. The transverse axis is the axis of the conic that passes through the two foci.