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  • Through two points <img alt="\mathrm{ M(c, 0) and M^{\prime}(-c, 0)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20M%28c%2C%200%29%20and%20M%5E%7B%5Cprime%7D%28-c%2C%200%29%7

Through two points \mathrm{ M(c, 0) and M^{\prime}(-c, 0)}, two chords \mathrm{P M Q \: and \: P^{\prime} M^{\prime} Q^{\prime}} are drawn to cut the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} in points \mathrm{P(a \cos \alpha, b \sin \alpha), Q(a \cos \beta, b \sin \beta), P^{\prime}(a \cos \gamma, b \sin \gamma), Q^{\prime}(a \cos \delta, b \sin \delta)}, then \mathrm{\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}\: \: and \: \: \frac{\delta}{2}=k}, where \mathrm{ k=}

 

Option: 1

-1


Option: 2

1


Option: 3

4


Option: 4

2


Answers (1)

best_answer

Using solved example (5),

\mathrm{\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{c-a}{c+a}}

\mathrm{and \: \tan \frac{\gamma}{2} \tan \frac{\delta}{2}=\frac{-c-a}{-c+a}}

On multiplying, we obtain

\mathrm{\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2} \tan \frac{\delta}{2}=1 \quad}

Hence option 2 is correct.
 

Posted by

shivangi.shekhar

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