15. Given that the vertex of a parabola lies on the $x$-axis, and the directrix is the $y$-axis. The shortest distance from point $\Lambda(4,0)$ to this parabola is 2, find the equation of this parabola.
15. Let the equation of the parabola be $y^{2}=2 p(x-a)$. From $a=\frac{p}{2}$, we get $2 p=4 a$, hence the equation is $y^{2}=4 a(x-a), x \in[a,+\infty)$.
Let $Q(x, y)$ be a moving point on the parabola, then
$$
\begin{aligned}
|A Q| & =\sqrt{(x-4)^{2}+y^{2}} \\
& =\sqrt{x^{2}-(8-4 a) x+16-4 a^{2}} \\
& =\sqrt{[x-(4-2 a)]^{2}+16 a-8 a^{2}} .
\end{aligned}
$$

When $4-2 a \geqslant a$, i.e., $a \leqslant \frac{4}{3}$, $|A Q|_{\text {min }}=\sqrt{16 a-8 a^{2}}=2$,
solving gives $a=1 \pm \frac{\sqrt{2}}{2}$. Since $a=1+\frac{\sqrt{2}}{2}>\frac{4}{3}$, it is discarded;
When $4-2 a < a$, i.e., $a > \frac{4}{3}$, $|A Q|_{\text{min}}=|a-4|=2$,
solving gives $a=2$ or $6$.
Therefore, the equations of the parabolas are
$$
\begin{array}{c}
y^{2}=(4-2 \sqrt{2})\left(x-1+\frac{\sqrt{2}}{2}\right) ; \\
y^{2}=8(x-2) ; \\
y^{2}=24(x-6) .
\end{array}
$$