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If any two chords be drawn through two points on the major axis of an ellipse equidistant from the center,show that $\tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\gamma}{2}\tan\frac{\delta}{2}=1$,where $\alpha,\beta,\gamma,\delta$ are the eccentric angles of the extremities of the chords.

I tried to solve it.Let the chords cut the major axis at $(-d,0)$ and $(d,0)$.

The equation of the chords are $y=\frac{-b}{a}\cot\frac{\alpha+\beta}{2}(x+d)$ and $y=\frac{-b}{a}\cot\frac{\gamma+\delta}{2}(x-d)$

and then i stuck.What should i do to reach the answer.Please help.

Brahmagupta
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    Let if $(a\cos\alpha, b\sin\alpha);(a\cos\gamma, b\sin\gamma); (d,0)$ forms a chord

    $$\dfrac{b\sin\alpha- b\sin\gamma}{a\cos\alpha-a\cos\gamma}=\dfrac{b\sin\alpha- d}{a\cos\alpha-0}$$

    $$d=b\left(\tan\dfrac{\alpha+\gamma}2+\sin\alpha\right)$$

    Similarly for $(a\cos\beta, b\sin\beta);(a\cos\delta, b\sin\delta); (d,0)$

    $$-d=b\left(\tan\dfrac{\beta+\delta}2+\sin\delta\right)$$

    $$\tan\dfrac{\alpha+\gamma}2+\sin\alpha+\tan\dfrac{\beta+\delta}2+\sin\delta=0$$

     – lab bhattacharjee Aug 26 '15 at 17:13
  • why don't produce your answer here? – lab bhattacharjee Aug 27 '15 at 07:56

1 Answers1

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The equation to the chord of the ellipse joining two points with eccentric angles $\alpha$ and $\beta$ is $\frac{x}{a}\cos \frac{\alpha+\beta}{2}+\frac{y}{b}\sin \frac{\alpha+\beta}{2}=\cos\frac{\alpha-\beta}{2}$

Similarly,The equation to the chord of the ellipse joining two points with eccentric angles $\gamma$ and $\delta$ is $\frac{x}{a}\cos \frac{\gamma+\delta}{2}+\frac{y}{b}\sin \frac{\gamma+\delta}{2}=\cos\frac{\gamma-\delta}{2}$

Since,first chord passes through $(d,0)$ and the second chord passes through $(-d,0)$.

$\frac{d}{a}\cos \frac{\alpha+\beta}{2}=\cos\frac{\alpha-\beta}{2}$ and

$\frac{-d}{a}\cos \frac{\gamma+\delta}{2}=\cos\frac{\gamma-\delta}{2}$

$\Rightarrow \frac{\cos\frac{\alpha-\beta}{2}}{\cos\frac{\alpha+\beta}{2}}=-\frac{\cos\frac{\gamma-\delta}{2}}{\cos\frac{\gamma+\delta}{2}}$

$\Rightarrow \frac{\cos\frac{\alpha}{2}\cos\frac{\beta}{2}+\sin\frac{\alpha}{2}\sin\frac{\beta}{2}}{\cos\frac{\alpha}{2}\cos\frac{\beta}{2}-\sin\frac{\alpha}{2}\sin\frac{\beta}{2}}=-\frac{\cos\frac{\gamma}{2}\cos\frac{\delta}{2}+\sin\frac{\gamma}{2}\sin\frac{\delta}{2}}{\cos\frac{\gamma}{2}\cos\frac{\delta}{2}-\sin\frac{\gamma}{2}\sin\frac{\delta}{2}}$

Now apply componendo and dividendo on both sides,we will get the answer.

Brahmagupta
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