Example 3. Derive the equation of an ellipse symmetric with respect to the coordinate axes, passing through the point $A(-3,1.75)$ and having an eccentricity $\varepsilon=0.75$.
Solution. We have a system of equations with respect to the parameters $a, b, c=\sqrt{a^{2}-b^{2}}:$

$$
\left\{\begin{array}{l}
\left.\frac{(-3)^{2}}{a^{2}}+\frac{1.75^{2}}{b^{2}}=1 \quad \text { (the ellipse passes through point } A\right) \\
\frac{\sqrt{a^{2}-b^{2}}}{a}=0.75, \quad \text { or } \frac{a^{2}-b^{2}}{a^{2}}=0.5625 \text { (the eccentricity is given). }
\end{array}\right.
$$

From the second equation, we find:

$$
1-\left(\frac{b}{a}\right)^{2}=0.5625, \quad \text { i.e. }\left(\frac{b}{a}\right)^{2}=0.4375, \quad \text { or } b^{2}=0.4375 a^{2} .
$$

Substituting this into the first equation, we get $\frac{9}{a^{2}}+\frac{7}{a^{2}}=1$, then $a^{2}=16, \quad b^{2}=7, \quad c^{2}=9$

The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$.