$\qquad$ $\star$ Find the equation of the parabola that is tangent to the $x$-axis at $(1,0)$ and tangent to the $y$-axis at $(0,2)$. Also, find the axis of symmetry and the coordinates of the vertex of the parabola.
Obviously, the parabola in question does not pass through the origin. Any conic section that does not pass through the origin has the equation $A x^{2}+B x y+C y^{2}+D x+E y+1=0$. Since the parabola is tangent to the $x$-axis at $(1,0)$, we have $A+$ $D+1=0$, (1) and $A x^{2}+D x+1=0$ has two equal roots, thus $D^{2}-4 A=0$. (2) Solving (1) and (2) gives $A=1, D=-2$.
Since the parabola is tangent to the $y$-axis at $(0,2)$, we have $C_{2}^{2}+E_{2}+1=0$.
(3) and $C y^{2}+E y+$ $1=0$ has two equal roots, thus $E^{2}-4 C=0$. (4) Solving (3) and (4) gives $C=\frac{1}{4}, E=-1$. Since it is a parabola, we have $B^{2}=4 A C$, giving $B= \pm 1$. Therefore, we obtain two equations
$$
\begin{array}{l}
x^{2}+x y+\frac{1}{4} y^{2}-2 x-y+1=0, \\
x^{2}-x y+\frac{1}{4} y^{2}-2 x-y+1=0,
\end{array}
$$

Since (5) can be transformed into $\left(x+\frac{1}{2} y-1\right)^{2}=0$, it is a line, so (6) is the required equation. Clearly, (6) can be transformed into
$$
4 x^{2}-4 x y+y^{2}-8 x-4 y+4=0 \text {. }
$$

Let $u=2 x-y, v=x+2 y$, then $x=\frac{2 u+v}{5}, y=\frac{-u+2 v}{5}$. Then (7) can be transformed into $u^{2}-\frac{12}{5} u-$ $\frac{16}{5} v+4=0$, i.e., $\left(u-\frac{6}{5}\right)^{2}-\frac{16}{5}\left(v-\frac{4}{5}\right)=0$.

Thus, this is a parabola with the axis of symmetry $u=\frac{6}{5}$, and the vertex at $\left(\frac{6}{5}, \frac{4}{5}\right)$, i.e., the equation of the axis of symmetry of the parabola is $2 x-y=\frac{6}{5}$, and the vertex coordinates are $\left(\frac{2 \times \frac{6}{5}+\frac{4}{5}}{5}, \frac{-\frac{6}{5}+2 \times \frac{4}{5}}{5}\right)=\left(\frac{16}{25}, \frac{2}{25}\right)$.