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  • If S and be 2 foci and P is any point on <img alt="\mathrm{\frac{

If S and \mathrm{S^{\prime}} be 2 foci and P is any point on \mathrm{\frac{x^2}{48}+\frac{y^2}{36}=1} such that \mathrm{\angle P S S^{\prime}=\theta, \angle P S^{\prime} S=\phi}, if the ratio of \mathrm{\frac{1-\cos \theta}{1+\cos \phi}} and \mathrm{\frac{1+\cos \theta}{1-\cos \phi}} is \mathrm{1: \lambda, then \, \lambda=}

Option: 1

7


Option: 2

8


Option: 3

9


Option: 4

5


Answers (1)

best_answer

Here, \mathrm{a=4 \sqrt{3}, b=6}

\mathrm{ \begin{aligned} & e=\sqrt{1-\frac{36}{48}}=\frac{1}{2} \\ & S S^{\prime}=2 a e=4 \sqrt{3} \end{aligned} }

Let \mathrm{P S=x \, \, and \, \, P S^{\prime}=y}

\mathrm{ \therefore \text { In } \Delta P S S^{\prime}, s=\frac{P S+P S^{\prime}+S S^{\prime}}{2} }

\mathrm{ =\frac{2 a+2 a e}{2}=a(1+e) }  ....\mathrm{ \text { (i) } \quad\left[\because s=\frac{a+b+c}{2}\right] }

\mathrm{\left[\because\right. In \, \, any \, \, ellipse, P S+P S^{\prime}=2 a\, ( if \left.a>b)\right] }

\mathrm{ \begin{aligned} & \text { Now, } \tan \frac{\theta}{2} \tan \frac{\phi}{2}=\sqrt{\frac{(s-x)(s-2 a e)}{s(s-y)}} \cdot \sqrt{\frac{(s-y)(s-2 a e)}{s(s-x)}} \\\\ & =\frac{s-2 a e}{s}=\frac{a(1+e)-2 a e}{a(1+e)}[\text { From (i)] } \\\\ & =\frac{1-e}{1+e}=\frac{1-1 / 2}{1+1 / 2}=\frac{1}{3} \\\\ & \Rightarrow \frac{2 \sin ^2(\theta / 2) \cdot 2 \sin ^2(\phi / 2)}{2 \cos ^2(\theta / 2) \cdot 2 \cos ^2(\phi / 2)}=\frac{1}{9} \\\\ & \Rightarrow \frac{(1-\cos \theta)\left(1-\cos ^2\right)}{(1+\cos \theta)(1+\cos \phi)}=\frac{1}{9} \\\\ & \therefore \frac{1-\cos \theta}{1+\cos \phi}: \frac{1+\cos \theta}{1-\cos \phi}=1: 9 \end{aligned} }

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Suraj Bhandari

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