Example 2. Given that an ellipse passes through four points $\mathrm{A}(-2,0), \mathrm{B}$ $(2,0), C(0,-1)$ and D $(0,1)$ and is tangent to $x-2 y=4$. Find the equation of this ellipse.
Solve the following four points
as the four vertices of a quadrilateral, the equations of these four sides are respectively 8
$$
\begin{array}{l}
\text { AD: } x-2 y+2=0, \\
\text { BC: } x-2 y-2=0, \\
\text { AC: } x+2 y+2=0, \\
\text { BD: } x+2 y-2=0 . \\
\therefore(x-2 y+2)(x-2 y-2)+m \\
\\
\cdot(x+2 y+2)(x+2 y-2)=0, \\
\quad(m+1) x^{2}+4(m-1) x y+4(m+
\end{array}
$$
$$
\text { 1) } y^{2}-8=0 \text {. }
$$

When it is an ellipse, it should satisfy
$$
\Delta=b^{2}-4 \mathrm{ac}0 \text {. }
$$

Since the curve sought is tangent to the line $x-2 y=4$, substituting the line into the equation of the curve yields $2 y^{2}-4 y+2 m$ $+1=0$.
To make $\triangle^{\prime}=b^{\prime}-4 a^{\prime} c^{\prime}$
$$
=16-8(2 m+1)=0 \text {, }
$$

thus $\mathrm{m}=\frac{1}{2}$.
From (1) and (2), we get $m=\frac{1}{2}$,
therefore, the equation of the ellipse sought is:
$$
3 x^{2}-4 x y+12 y^{2}-16=0
$$