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  • If tangent and normal at P to ellipse <img alt="\mathrm{\frac{x^2}{25}+\frac{y^2}{16}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7B25%7D&plus;%

If tangent and normal at P to ellipse \mathrm{\frac{x^2}{25}+\frac{y^2}{16}=1}  intersect major axis at T and N respectively in such a way that ratio of area of  \mathrm{\triangle P T N \text { and } \triangle P S S^{\prime} \text { is } \frac{91}{60}} , then area of \mathrm{\triangle P S S^{\prime}}  is (S and S′ are foci)

 

Option: 1

6 \sqrt{3} \text { sq. units }


Option: 2

12 \sqrt{3} \text { sq. units }


Option: 3

4 \sqrt{3} \text { sq. units }


Option: 4

3 \sqrt{3} \text { sq. units }


Answers (1)

best_answer

\mathrm{\begin{aligned} &e=\sqrt{1-\frac{b^2}{a^2}}=\frac{3}{5}\\ &T \equiv\left(\frac{a}{\cos \theta}, 0\right) \quad N=\left(\frac{a^2-b^2}{a \sec \theta}, 0\right) \end{aligned}}

\mathrm{ \begin{gathered} \frac{a \sec \theta-a e^2 \cos \theta}{2 a e}=\frac{91}{60} \Rightarrow \quad \cos \theta=\frac{1}{2}, \sin \theta=\frac{\sqrt{3}}{2} \\ \Delta P S S^{\prime}=\frac{1}{2} \times 6 \times 4 \times \frac{\sqrt{3}}{2}=6 \sqrt{3} \end{gathered} sq. units}

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rishi.raj

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