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Question: \(\left( a\sec\theta,b\tan\theta \right)\)and \(\left( a\sec\varphi,b\tan\varphi \right)\) are the e...

(asec⁡θ,btan⁡θ)\left( a\sec\theta,b\tan\theta \right)and (asec⁡φ,btan⁡φ)\left( a\sec\varphi,b\tan\varphi \right) are the ends of a focal chord of x2a2−y2b2=1\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, then tan⁡φ2\tan\frac{\varphi}{2} equals to

A

e−1e+1\frac{e - 1}{e + 1}

B

1−e1+e\frac{1 - e}{1 + e}

C

1+e1−e\frac{1 + e}{1 - e}

D

e+1e−1\frac{e + 1}{e - 1}

Answer

1−e1+e\frac{1 - e}{1 + e}

Explanation

Solution

Equation PQ is

xacos⁡(θ−φ2)−ybsin⁡(θ+φ2)=cos⁡(θ+φ2)\frac{x}{a}\cos\left( \frac{\theta - \varphi}{2} \right) - \frac{y}{b}\sin\left( \frac{\theta + \varphi}{2} \right) = \cos\left( \frac{\theta + \varphi}{2} \right)

Its passes through (ae, 0)

∴\therefore ecos⁡(θ−φ2)−0=cos⁡(θ+φ2)e\cos\left( \frac{\theta - \varphi}{2} \right) - 0 = \cos\left( \frac{\theta + \varphi}{2} \right)⇒

cos⁡(θ−φ2)cos⁡(θ+φ2)=1e\frac{\cos\left( \frac{\theta - \varphi}{2} \right)}{\cos\left( \frac{\theta + \varphi}{2} \right)} = \frac{1}{e}

cos⁡(θ−φ2)−cos⁡(θ+φ2)cos⁡(θ−φ2)+cos⁡(θ+φ2)=1−e1+e\frac{\cos\left( \frac{\theta - \varphi}{2} \right) - \cos\left( \frac{\theta + \varphi}{2} \right)}{\cos\left( \frac{\theta - \varphi}{2} \right) + \cos\left( \frac{\theta + \varphi}{2} \right)} = \frac{1 - e}{1 + e}

(By componentdo & dividendo method)

⇒ 1−e1+e\frac{1 - e}{1 + e}