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Question: If (a secθ, b tanθ) and (a secφ, b tanφ) are the ends of a focal chord of \(\frac{x^{2}}{a^{2}} - \f...

If (a secθ, b tanθ) and (a secφ, b tanφ) are the ends of a focal chord of x2a2−y2b2\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, then tan θ2 tan φ2\frac{\theta}{2}\text{ 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 of the chord connecting the points (a sec θ, b tan θ) and (a sec φ, b tan φ) 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)If it passes through (ae, 0), we have, e cos(θ−φ2)\left( \frac{\theta - \varphi}{2} \right)= cos (θ+φ2)\left( \frac{\theta + \varphi}{2} \right)

⇒ e = cos⁡(θ+φ2)cos⁡(θ−φ2)\frac{\cos\left( \frac{\theta + \varphi}{2} \right)}{\cos\left( \frac{\theta - \varphi}{2} \right)} = 1−tan⁡θ2.tan⁡φ21+tan⁡θ2.tan⁡φ2\frac{1 - \tan\frac{\theta}{2}.\tan\frac{\varphi}{2}}{1 + \tan\frac{\theta}{2}.\tan\frac{\varphi}{2}}

⇒ tan⁡θ2.tan⁡φ2=1−e1+e\tan\frac{\theta}{2}.\tan\frac{\varphi}{2} = \frac{1 - e}{1 + e}.