Example 3 Given
$$
\left\{\begin{array}{l}
(a-b) \sin \frac{\theta}{2} \cos \frac{\varphi}{2}+(a+b) \cos \frac{\theta}{2} \sin \frac{\varphi}{2}=0(a, b>0) \\
\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1 \\
\frac{x}{a} \cos \varphi+\frac{y}{b} \sin \varphi=1
\end{array}\right.
$$

Try to eliminate $\theta, \varphi$, and establish the identity between $x, y, a, b$.
Solve: From (1) we get $b \sin \frac{\theta-\varphi}{2}=a \sin \frac{\theta+\varphi}{2}$
(2)(3) Eliminating $\frac{y}{b}$ yields $\frac{x}{a}=\frac{\sin \theta-\sin \varphi}{\sin (\theta-\varphi)}=\frac{\cos \frac{\theta+\varphi}{2}}{\cos \frac{\theta-\varphi}{2}}$, thus $x \cos \frac{\theta-\varphi}{2}=a \cos \frac{\theta+\varphi}{2}$

While eliminating $\frac{x}{a}$ yields $y \cos \frac{\theta-\varphi}{2}=b \sin \frac{\theta+\varphi}{2}$
(4) $\div$ (6) gives $\tan \frac{\theta-\varphi}{2}=\frac{a y}{b^{2}}$

Also, $(5)^{2} \cdot b^{2}+(6)^{2} \cdot a^{2}$ gives $\left(b^{2} x^{2}+a^{2} y^{2}\right) \cos ^{2} \frac{\theta-\varphi}{2}=a^{2} b^{2}$, from (7) we get $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=$ $\sec ^{2} \frac{\theta-\varphi}{2}=1+\tan ^{2} \frac{\theta-\varphi}{2}=1+\frac{a^{2} y^{2}}{b^{4}}$, which is $\frac{x^{2}}{a^{2}}+\frac{b^{2}-a^{2}}{b^{4}} y^{2}=1$.