The parabola with equation $y=-\frac{1}{4} x^{2}+5 x-21$ has its vertex at point $A$ and crosses the $x$-axis at $B(b, 0)$ and $F(f, 0)$ where $b<f$. A second parabola with its vertex at $B$ passes through $A$ and crosses the $y$-axis at $D$. What are the coordinates of $D$ ?
By completing the square, we have $y=-\frac{1}{4}(x-10)^{2}+4$, which means the vertex of the parabola is at $A(10,4)$.

If we set $y=0$, we have $-\frac{1}{4}(x-10)^{2}+4=0$ or $-\frac{1}{4}(x-10)^{2}=-4$, so $(x-10)^{2}=16$.

Taking square roots of both sides, we have $x-10= \pm 4$ or $x=10 \pm 4$.

Therefore, the parabola passes through $(6,0)$ and $(14,0)$.

From the information given, this means the coordinates of $B$ are $(6,0)$ and the coordinates of $F$ are $(14,0)$.

The second parabola has its vertex at $B(6,0)$, so its equation is $y=a(x-6)^{2}$ for some real number $a$.

Using the fact that it passes through $A(10,4)$, we get $4=a(10-6)^{2}=16 a$, or $a=\frac{1}{4}$.

Therefore, the equation of the second parabola is $y=\frac{1}{4}(x-6)^{2}$.

Substituting $x=0$, we get $y=\frac{1}{4}(-6)^{2}=\frac{36}{4}=9$.

The second parabola crosses the $y$-axis at $D(0,9)$.

ANSWER: $(0,9)$