Given, equation $\frac{1}{r}=\frac{1}{8}+\frac{3}{8} \cos \theta$
$\frac{1}{r}=\frac{1}{8}+\frac{3}{8} \cos \theta$
$\Rightarrow \frac{1}{r}=\frac{1+3 \cos \theta}{8}$
$\Rightarrow \frac{8}{r}=1+3 \cos \theta$
Compairing with $\frac{1}{r}=1+e \cos \theta$
we get $e=3$
Hence essentsieily is $e=3$ which greater than $1$
$\therefore $ The given conic is hyperbola