jmc

Dividing by 16 gives the standard form of the equation of this ellipse, $$\frac{x^2}{(\frac{16}{25})}+\frac{y^2}{4}=1.$$Then, the semimajor and semiminor axes have lengths $\sqrt{4}=2$ and $\sqrt{\frac{16}{25}}=\frac{4}{5}$. By the area formula for an ellipse, the area of the ellipse is $2\cdot \frac{4}{5}\cdot \pi =\boxed{\frac{8\pi }{5}}.$