In the coordinate plane, let $F = (4,0).$  Let $P$ be a point, and let $Q$ be the projection of the point $P$ onto the line $x = \frac{25}{4}.$  The point $P$ traces a curve in the plane, so that
\[\frac{PF}{PQ} = \frac{4}{5}\]for all points $P$ on the curve.  Find the area of the region formed by the curve.

[asy]
unitsize(1 cm);

pair P, F, Q;

F = (4,0);
P = (5*Cos(60),3*Sin(60));
Q = (25/4,3*Sin(60));

draw(F--P--Q);
draw((25/4,-1)--(25/4,3),dashed);

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

label("$x = \frac{25}{4}$", (25/4,-1), S);
[/asy]
Let $P = (x,y)$; then $Q = \left( \frac{25}{4}, y \right).$  The condition $\frac{PF}{PQ} = \frac{4}{5}$ becomes
\[\frac{\sqrt{(x - 4)^2  +y^2}}{|\frac{25}{4} - x|} = \frac{4}{5}.\]Hence, $\sqrt{(x - 4)^2 + y^2} = \left| 5 - \frac{4}{5} x \right|,$ or
\[5 \sqrt{(x - 4)^2 + y^2} = |25 - 4x|.\]Squaring both sides, we get
\[25 ((x - 4)^2 + y^2) = (25 - 4x)^2.\]This simplifies to $9x^2 + 25y^2 = 225,$ or
\[\frac{x^2}{25} + \frac{y^2}{9} = 1.\]Hence, the curve is an ellipse, and its area is $\pi \cdot 5 \cdot 3 = \boxed{15 \pi}.$