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  • S and S are foci of   <img alt="\mathrm {\frac{x^2}{25}+\frac{y^2}{9}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%20%7B%5Cfrac%7Bx%5E2%7D%7B25%7D&plus

S and S are foci of   \mathrm {\frac{x^2}{25}+\frac{y^2}{9}=1}  and  \mathrm P\left(\frac{\pi}{3}\right)  is a point on this ellipse such that incentre of  \mathrm {\triangle S P S^{\prime}}  is \mathrm {\left(\alpha, \frac{\alpha}{\sqrt{\alpha^2-1}}\right)} then
 

Option: 1

2


Option: 2

3


Option: 3

\sqrt{2}


Option: 4

none of these


Answers (1)

best_answer

\begin{aligned} & \text { } e=\sqrt{1-\frac{9}{25}}=\frac{4}{5} \\ & \therefore \mathrm {\quad S(4,0), S^{\prime}(-4,0) }\\ & \mathrm {p\left(\frac{\pi}{3}\right) \text { corresponds to }\left(5 \cos \frac{\pi}{3}, 3 \sin \frac{\pi}{3}\right) \text { i.e., }\left(\frac{5}{2}, \frac{3 \sqrt{3}}{2}\right)} \\ &\mathrm { P S=3, P S^{\prime}=7\: \: \&\: \: S S^{\prime}=8} \end{aligned}

therefore Incentre is given by

\begin{aligned} & \left\{\frac{8\left(\frac{5}{2}\right)+3(-4)+7(4)}{8+3+7}, \frac{8\left(\frac{3 \sqrt{3}}{2}\right)+3(0)+7(0)}{8+3+7}\right\} \\ & \text { i.e., }\left(2, \frac{2}{\sqrt{3}}\right) \text { i.e., }\left(2, \frac{2}{\sqrt{2^2-1}}\right) \Rightarrow \alpha=2 \end{aligned}

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jitender.kumar

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