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