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  • Let C be the largest circle centred at (2,0) and inscribed in the ellipse<img alt="\frac{x^2}{36}+\frac{y^2}{16}=1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bx%5E2%7D%7B36%7Da

Let C be the largest circle centred at (2,0) and inscribed in the ellipse\frac{x^2}{36}+\frac{y^2}{16}=1  If (1, α) lies on C, then \mathrm{10}\alpha^{2}  is equal to 

Option: 1

118


Option: 2

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Option: 3

--


Option: 4

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Answers (1)

best_answer

E: \frac{x^2}{36}+\frac{y^2}{16}=1 \& C:(x-2)^2+y^2=r^2

For largest circle r is maximum
\begin{aligned} & \mathrm{P}(6 \cos \theta, 4 \sin \theta) \\ & N_{\mathrm{P}}: 6 \times \sec \theta-4 \mathrm{y} \operatorname{cosec} \theta=20 \text { pass }(2,0) \\ & 12 \sec \theta=20 \Rightarrow \cos \theta=\frac{3}{5} \\ & \text\! { Now P: }\left(6 \times \frac{3}{5}, 4 \times \frac{4}{5}\right) \Rightarrow P:\left(\frac{18}{5}, \frac{16}{5}\right) \end{aligned}

\begin{aligned} & r=\sqrt{\left(2-\frac{18}{5}\right)^2+\left(\frac{16}{5}\right)^2} \\ & r=\frac{\sqrt{64+256}}{5}=\frac{8 \sqrt{5}}{5}=\frac{8}{\sqrt{5}} \\ & \mathrm{C}:=(\mathrm{x}-2)^2+y^2=\frac{64}{5} \\ & \text \!{ Now }(1, \alpha) \text { lies on } \mathrm{C} \\ & \Rightarrow(1-2)^2+\alpha^2=\frac{64}{5} \\ & \alpha^2=\frac{64}{5}-1 \\ & \alpha^2=\frac{59}{5} \Rightarrow 10 \alpha^2=118 \end{aligned}

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Irshad Anwar

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