Q98 / 160 · TS EAMCET 2004 · Mathematics
The eccentricity of the conic $36 x^2+144 y^2-36 x-96 y-119=0$ is
- Correct answerA$\frac{\sqrt{3}}{2}$
- B$\frac{1}{2}$
- C$\frac{\sqrt{3}}{4}$
- D$\frac{1}{\sqrt{3}}$
Solution
Given equation of conic is
$$
\begin{aligned}
& 36 x^2+144 y^2-36 x-96 y-119=0 \\
& \Rightarrow \quad 36\left(x^2-x\right)+144\left(y^2-\frac{2}{3}\right)=119 \\
& \Rightarrow \quad 36\left(x^2-x+\frac{1}{4}\right)+144\left(y^2-\frac{2}{3}+\frac{1}{9}\right) \\
& =119+9+16 \\
& \Rightarrow \quad 36\left(x-\frac{1}{2}\right)^2+144\left(y-\frac{1}{3}\right)^2=114 \\
& \Rightarrow \quad \frac{\left(x-\frac{1}{2}\right)^2}{4}+\frac{\left(y-\frac{1}{3}\right)^2}{1}=1 \\
&
\end{aligned}
$$
This is the equation of ellipse.
Here,
$$
\therefore \quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}
$$
Hence, the correct answer is option: 'A'.