Question

the eccentricity of the conic 36
${x}^{2} + 144 {y }^{2} - 36x - 96y - 119 = 0$
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Answer 1

$\textbf{Given:}$

$36x^2+144y^2-36x-96y-119=0$

$36(x^2-x)+144(y^2-\frac{2}{3}y)-119=0$

$36(x^2-x+\frac{1}{4}-\frac{1}{4})+144(y^2-\frac{2}{3}y+\frac{1}{9}-\frac{1}{9})-119=0$

$36(x-\frac{1}{2})^2-9+144(y-\frac{1}{3})^2-16-119=0$

$36(x-\frac{1}{2})^2+144(y-\frac{1}{3})^2=144$

$\text{Divide both sides by 144}$

$\displaystyle\frac{(x-\frac{1}{2})^2}{4}+\frac{(y-\frac{1}{3})^2}{1}=1$

$\text{Clealry this is equation of ellipse with major axis is parallel to x axis}$

$\text{Here, }a^2=4\;\text{and}\;b^2=1$

$\text{Eccentricity}$

$=\displastyle\sqrt{1-\frac{b^2}{a^2}}$

$=\displastyle\sqrt{1-\frac{1}{4}}$

$=\displastyle\sqrt{\frac{3}{4}}$

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

$\therefore\text{The eccentricity is}\;\frac{\sqrt{3}}{2}$