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  • Consider the ellipse  and a point P on it in the first

Consider the ellipse \mathrm{x^2+3 y^2=6} and a point P on it in the first quadrant at a distance of 2 units from the centre. Find the eccentric angle of P.

Option: 1

\mathrm{\frac{\pi}{6}}


Option: 2

\mathrm{\frac{\pi}{4}}


Option: 3

\mathrm{\frac{\pi}{3}}


Option: 4

\mathrm{\frac{\pi}{2}}


Answers (1)

best_answer

Equation of ellipse is \mathrm{x^2+3 y^2=6}
Equation of auxiliary circle is \mathrm{x^2+y^2=6}

Since \mathrm{P \equiv\left(x_1, y_1\right) \& Q \equiv\left(x_1, y_2\right)} lie on the ellipse and the circle respectively we have,

\mathrm{ \begin{aligned} & x_1{ }^2+3 y_1{ }^2=6 \, \, \, \, \, ....(1)\\ & x_1{ }^2+y_2{ }^2=6\, \, \, \, \, \, ....(2) \end{aligned} }

\mathrm{\therefore 3 \mathrm{y}_1{ }^2-\mathrm{y}_2{ }^2=0 \Rightarrow \frac{\mathrm{y}_2}{\mathrm{y}_1}=\sqrt{3}}

\mathrm{\text { Again OP }=2 \Rightarrow x_1^2+y_1^2=4\, \, \, \, \, \, \, ....(3)}
By (1) -(3), we get,

\mathrm{ 2 \mathrm{y}_1{ }^2=2 \Rightarrow \mathrm{y}_1{ }^2=1 }

\mathrm{\Rightarrow \mathrm{y}_1=1 } [P is in the first quadrant ]

\mathrm{ \therefore \mathrm{y}_2=\sqrt{3} }

Putting \mathrm{\mathrm{y}_1} in (1), we get

\mathrm{ x_1^2=3 \Rightarrow x_1=\sqrt{3} }

\mathrm{\therefore Eccentric \, \, angle \, \, of \, \mathrm{P}=\theta= \tan ^{-1}\left(\frac{\mathrm{y}_2}{\mathrm{x}_1}\right)=\tan ^{-1}\left(\frac{\sqrt{3}}{\sqrt{3}}\right)=\frac{\pi}{4} }

Posted by

Ajit Kumar Dubey

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