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Question: If α and β are eccentric angles of the ends of a focal chord of the ellipse \(\frac{x^{2}}{a^{2}} + ...

If α and β are eccentric angles of the ends of a focal chord of the ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, then tanα2\frac{\alpha}{2} tan β2\frac{\beta}{2} is equal to

A

1−e1+e\frac{1 - e}{1 + e}

B

e−1e+1\frac{e - 1}{e + 1}

C

e+1e−1\frac{e + 1}{e - 1}

D

None of these

Answer

e−1e+1\frac{e - 1}{e + 1}

Explanation

Solution

The coordinates of the end points of the focal chord are (a cos α. b sin α) and (a cos β, b sin β). Therefore the equation of focal chord is

xacos⁡(α+β2)=cos⁡(α−β2)\frac{x}{a}\cos\left( \frac{\alpha + \beta}{2} \right) = \cos\left( \frac{\alpha - \beta}{2} \right)This passes through (ae, 0)

∴ aeα\frac{ae}{\alpha} cos (α+β2)=cos⁡(α−β2)\left( \frac{\alpha + \beta}{2} \right) = \cos\left( \frac{\alpha - \beta}{2} \right)

⇒ cos⁡(α+β2)cos⁡(α−β2)=1e\frac{\cos\left( \frac{\alpha + \beta}{2} \right)}{\cos\left( \frac{\alpha - \beta}{2} \right)} = \frac{1}{e}

⇒cos⁡(α+β2)−cos⁡(α−β2)cos⁡(α+β2)+cos⁡(α−β2)=1−e1+e\frac{\cos\left( \frac{\alpha + \beta}{2} \right) - \cos\left( \frac{\alpha - \beta}{2} \right)}{\cos\left( \frac{\alpha + \beta}{2} \right) + \cos\left( \frac{\alpha - \beta}{2} \right)} = \frac{1 - e}{1 + e}

⇒ −2sin⁡α2sin⁡β22cos⁡α2cos⁡β2=1−e1+e- \frac{2\sin\frac{\alpha}{2}\sin\frac{\beta}{2}}{2\cos\frac{\alpha}{2}\cos\frac{\beta}{2}} = \frac{1 - e}{1 + e}

⇒ - tan α2tan⁡β2=1−e1+e\frac{\alpha}{2}\tan\frac{\beta}{2} = \frac{1 - e}{1 + e} ⇒ tan⁡α2tan⁡β2=e−1e+1\tan\frac{\alpha}{2}\tan\frac{\beta}{2} = \frac{e - 1}{e + 1}