The eccentricity of the conic 36x^2 + 144y^2 | vedclass

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Question

(A) The given equation is $36x^2 + 144y^2 - 36x - 96y - 119 = 0$. Rearranging the terms,we get $36(x^2 - x) + 144(y^2 - \frac{2}{3}y) = 119$. Completi

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(A) The given equation is $36x^2 + 144y^2 - 36x - 96y - 119 = 0$. Rearranging the terms,we get $36(x^2 - x) + 144(y^2 - \frac{2}{3}y) = 119$. Completing the square,we have $36(x^2 - x + \frac{1}{4}) + 144(y^2 - \frac{2}{3}y + \frac{1}{9}) = 119 + 9 + 16$. This simplifies to $36(x - \frac{1}{2})^2 + 144(y - \frac{1}{3})^2 = 144$. Dividing by $144$,we get $\frac{(x - 1/2)^2}{4} + \frac{(y - 1/3)^2}{1} = 1$. This is an ellipse with $a^2 = 4$ and $b^2 = 1$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

The eccentricity of the conic $36x^2 + 144y^2 - 36x - 96y - 119 = 0$ is

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