Given curve,
$b y^2=(x+a)^3$
On differentiating w.r.t. $x$, we get
$2 b y \frac{d y}{d x}=3(x+a)^2$
$\begin{aligned} & \text { Length of subnormal }=y \frac{d y}{d x}=y\left(\frac{3(x+a)^2}{2 b y}\right) \\ & \text { Length of subtanget }=y \frac{d x}{d y}=y\left(\frac{2 b y}{3(x+a)^2}\right)\end{aligned}$
Given, $P($ Subnormal length $)=q($ subtangent length) ${ }^2$
$\Rightarrow \quad P\left(y \frac{d y}{d x}\right)=q\left(y \frac{d x}{d y}\right)^2 \Rightarrow \frac{P}{q}=y\left(\frac{d x}{d y}\right)^3$
$\Rightarrow \quad \frac{P}{q}=y\left(\frac{2 b y}{3(x+a)^2}\right)^3=\frac{8 y^4}{27} \frac{b^3}{\left((x+a)^3\right]^2}$
$\Rightarrow \quad \frac{P}{q}=\frac{8 b}{27}\left(\frac{b y^2}{(x+a)^3}\right)^2$
$\Rightarrow \quad \frac{P}{q}=\frac{8 b}{27}$ $\left[\because b y^2=(x+a)^3\right]$