In the coordinate plane, let $F = (5,0).$  Let $P$ be a point, and let $Q$ be the projection of the point $P$ onto the line $x = \frac{16}{5}.$  The point $P$ traces a curve in the plane, so that
\[\frac{PF}{PQ} = \frac{5}{4}\]for all points $P$ on the curve.  Find the equation of this curve.  (Enter it in standard form.)

[asy]
unitsize(1 cm);

pair P, F, Q;

F = (5,0);
P = (6,3*sqrt(5)/2);
Q = (16/5,3*sqrt(5)/2);

draw(F--P--Q);
draw((16/5,-1)--(16/5,4),dashed);

dot("$F$", F, S);
dot("$P$", P, NE);
dot("$Q$", Q, W);

label("$x = \frac{16}{5}$", (16/5,-1), S);
[/asy]
Let $P = (x,y).$  Then $Q = \left( \frac{16}{5}, y \right),$ so the equation $\frac{PF}{PQ} = \frac{5}{4}$ becomes
\[\frac{\sqrt{(x - 5)^2 + y^2}}{\left| x - \frac{16}{5} \right|} = \frac{5}{4}.\]Then $\sqrt{(x - 5)^2 + y^2} = \left| \frac{5}{4} x - 4 \right|,$ so
\[4 \sqrt{(x - 5)^2 + y^2} = |5x - 16|.\]Squaring both sides, we get
\[16x^2 - 160x + 16y^2 + 400 = 25x^2 - 160x + 256.\]This simplifies to
\[9x^2 - 16y^2 = 144,\]so
\[\boxed{\frac{x^2}{16} - \frac{y^2}{9} = 1}.\]Thus, the curve is a hyperbola.