Q98 / 160 · TS EAMCET 2009 · Mathematics
The eccentricity of the conic $\frac{5}{r}=2+3 \cos \theta+4 \sin \theta$ is
- A$\frac{1}{2}$
- B1
- C$\frac{3}{2}$
- Correct answerD$\frac{5}{2}$
Solution
Given,
$\begin{array}{ll}
\text { Given, } \quad \frac{5}{r}=2+3 \cos \theta+4 \sin \theta \\
\Rightarrow \quad \frac{5}{r}=2+5\left(\frac{3}{5} \cos \theta+\frac{4}{5} \sin \theta\right) \\
\Rightarrow \quad \frac{5 / 2}{r} & =1+\frac{5}{2}(\cos \phi \cos \theta+\sin \phi \sin \theta)
\end{array}$
$\left(\text { put } \cos \phi=\frac{3}{5} \text {, then } \sin \phi=\frac{4}{5}\right)$
$\Rightarrow \quad \frac{5 / 2}{r}=1+\frac{5}{2} \cos (\theta-\phi)$
It is of the form $\frac{l}{r}=1+e \cos \theta$
$\therefore \quad e=\frac{5}{2}$
Hence, the correct answer is option: 'D'.