8. Given that $a$, $b$, $c$, and $d$ are four positive constants, when the real numbers $x$, $y$ satisfy $a x^{2}+b y^{2}=1$, the minimum value of $c x+d y^{2}$ is $\qquad$ .
8. $-\frac{c}{\sqrt{a}}$.

Let $k=c x+d y^{2}$.
From $a x^{2}+b y^{2}=1$, we get $y^{2}=\frac{1-a x^{2}}{b}$.
Then $k=c x+\frac{d-a d x^{2}}{b}$
$$
=-\frac{a d}{b}\left(x-\frac{b c}{2 a d}\right)^{2}+\frac{d}{b}+\frac{b c^{2}}{4 a d} \text {. }
$$

From $a x^{2}+b y^{2}=1 \Rightarrow a x^{2} \leqslant 1$
$$
\Rightarrow-\frac{1}{\sqrt{a}} \leqslant x \leqslant \frac{1}{\sqrt{a}} \text {. }
$$

Notice that, $k$ is a quadratic function of $x$, its graph opens downwards, and the axis of symmetry is $x=\frac{b c}{2 a d}>0$. Therefore, when $x=-\frac{1}{\sqrt{a}}$, $y=0$, $k_{\min }=-\frac{c}{\sqrt{a}}$.