13.6. 7 * Let the axis of symmetry of the parabola be $2 x+y-1=0$, the directrix be $x-2 y-5=0$, and it is tangent to the line $2 y+3=0$. Find the equation of this parabola.
Let the focus of the parabola be $F(u, v)$, and it lies on the axis of symmetry, so $v=1-2 u$. According to the definition of a parabola, let $P(x, y)$ be a point on the parabola, then $(x-u)^{2}+(y-1+2 u)^{2}=\frac{(x-2 y-5)^{2}}{5}$, rearranging gives
$$
4 x^{2}+4 x y+y^{2}-10(u-1) x-10(3-2 u) y+25 u^{2}-20 u-20=0,
$$

Since the line $2 y+3=0$ is tangent to the parabola, substituting $y=-\frac{3}{2}$ into equation (1) yields a quadratic equation in $x$
$$
4 x^{2}+(4-10 u) x+25 u^{2}-50 u+\frac{109}{4}=0,
$$

which has equal roots, so $\Delta=(4-10 u)^{2}-16\left(25 u^{2}-50 u+\frac{109}{4}\right)=0$, solving this gives $u=\frac{7}{5}$ or $u=1$.

When $u=\frac{7}{5}$, equation (1) becomes $4 x^{2}+4 x y+y^{2}-4 x-2 y+1=0$, which is $(2 x+y-1)^{2}=0$, representing two coincident lines, which contradicts the problem statement and should be discarded.
When $u=1$, the required equation of the parabola is $4 x^{2}+4 x y+y^{2}-10 y-15=0$.