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If tangent and normal to  \mathrm {x^2+4 y^2=9}   at   \mathrm {\theta=\frac{\pi }{4} \: \: be \: \: x+\alpha y=\beta} and  \mathrm {\alpha x-y=\gamma}  respectively, then which of the following are in G.P.
 

Option: 1

\alpha, \beta, \gamma


Option: 2

\alpha, \sqrt{2} \beta, 2 \gamma


Option: 3

\alpha, \beta, 2 \sqrt{2} \gamma


Option: 4

 None of these


Answers (1)

best_answer

\because \mathrm {At \: \: \theta}  means at   \mathrm {(a \cos \theta, b \sin \theta)\: \: for \: \: ellipse\: \: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}


Given ellipse can be written as  \mathrm {\frac{x^2}{9}+\frac{y^2}{9 / 4}=1} 


\mathrm {\therefore At \: \: \theta=\frac{\pi}{4}\: \: on\: \: it \: \: means\: \: at \left(\frac{3}{\sqrt{2}}, \frac{3}{2 \sqrt{2}}\right)} 
Tangent at  \theta=\frac{\pi}{4}  is


\mathrm {x \cdot\left(\frac{3}{\sqrt{2}}\right)+4 y\left(\frac{3}{2 \sqrt{2}}\right)=9 \Rightarrow x+2 y=3 \sqrt{2}} 


Normal at   \theta=\frac{\pi}{4}  is


\mathrm {2 x-y=2\left(\frac{3}{\sqrt{2}}\right)-\frac{3}{2 \sqrt{2}}=\frac{9}{2 \sqrt{2}}}

\therefore \quad On comparing with given condition, we get

\begin{aligned} &\mathrm{ \alpha=2, \beta=3 \sqrt{2}, \gamma=\frac{9}{2 \sqrt{2}} }\\ \because \quad & \alpha \cdot 2 \sqrt{2} \gamma=2 \cdot 2 \sqrt{2} \cdot \frac{9}{2 \sqrt{2}}=18=(3 \sqrt{2})^2=\beta^2 \\ \therefore \quad & \alpha, \beta, 2 \sqrt{2} \gamma \text { are in G.P. } \end{aligned}
 

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Rakesh

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