Let $A\left(r_1\cos \theta , r_1\sin \theta \right)$

$A$ Lies on straight line $4x+3y=10$

$\Rightarrow 4\left(r_1\cos \theta \right)+3\left(r_1\sin \theta \right)=10$

$\Rightarrow r_1\left(4\cos \theta +3\sin \theta \right)=10$
$\Rightarrow r_1=\frac{10}{4\cos \theta +3\sin \theta }$

Let $B\left(-r_2\cos \theta , -r_2\sin \theta \right)$

$B$ lies on line $8x+6y+5=0$

$\Rightarrow 8\left(-r_2\cos \theta \right)+6\left(-r_2\sin \theta \right)+5=0$

$\Rightarrow r_2\left(8\cos \theta +6\sin \theta \right)-5=0$$\Rightarrow r_2=\frac{5}{8\cos \theta +6\sin \theta }$
$\therefore \frac{r_1}{r_2}=\frac{4}{1}$