4. If real numbers $x, y$ satisfy $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, then the minimum value of $\frac{x}{c^{2}}+\frac{y^{2}}{d^{2}}$ is ( ).
(A) $-\frac{b^{2}}{d^{2}}$
(B) $\frac{b^{2}}{d^{2}}$
(C) $\frac{|a|}{c^{2}}$
(D) $-\frac{|a|}{c^{2}}$
4. D.

From $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get $-|a| \leqslant x \leqslant|a|$.
Then $\frac{x}{c^{2}}+\frac{y^{2}}{d^{2}}=-\frac{b^{2}}{a^{2} d^{2}} x^{2}+\frac{x}{c^{2}}+\frac{b^{2}}{d^{2}}$.
Notice that, the above expression is a quadratic function of $x$, whose graph opens downwards and the axis of symmetry is greater than 0.

Therefore, when $x=-|a|$, the function value reaches its minimum, at this time, the original expression is $-\frac{|a|}{c^{2}}$.