Question

The eccentricity of the conic $\frac{5}{r}=2+3\cos \theta +4\sin \theta$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{3}{2}$
• D. $\frac{5}{2}$

Answer 1

Answer: (D)

Given, $\frac{5}{r}=2+3\cos \theta +4\sin \theta$
$\Rightarrow \frac{5}{r}=2+5\left(\frac{3}{5}\cos \theta +\frac{4}{5}\sin \theta \right)$
$\Rightarrow \frac{5/2}{r}=1+\frac{5}{2}(\cos \varphi \cos \theta +\sin \varphi \sin \theta )$
(put $\cos \varphi =\frac{3}{5}$, then $\sin \varphi =\frac{4}{5})$
$\Rightarrow \frac{5/2}{r}=1+\frac{5}{2}\cos (\theta -\varphi )$
It is of the form $\frac{l}{r}=1+e\cos \theta$
$\therefore e=\frac{5}{2}$