1. Given real numbers $x, y$ satisfy
$$
17\left(x^{2}+y^{2}\right)-30 x y-16=0 \text {. }
$$

then the maximum value of $f(x, y)=\sqrt{16 x^{2}+4 y^{2}-16 x y-12 x+6 y+9}$ is . $\qquad$
$-1.7$.
From $17\left(x^{2}+y^{2}\right)-30 x y-16=0$, we get
$$
\begin{array}{l}
(x+y)^{2}+16(x-y)^{2}=16 . \\
\text { Let }\left\{\begin{array}{l}
x+y=4 \cos \theta, \\
x-y=\sin \theta
\end{array}(\theta \in \mathbf{R})\right. \\
\Rightarrow\left\{\begin{array}{l}
x=2 \cos \theta+\frac{1}{2} \sin \theta, \\
y=2 \cos \theta-\frac{1}{2} \sin \theta .
\end{array}\right.
\end{array}
$$

Then $f(x, y)$
$$
=\sqrt{(3 \sin \theta+4 \cos \theta)^{2}-3(3 \sin \theta+4 \cos \theta)+9} \text {. }
$$

Therefore, when $3 \sin \theta+4 \cos \theta=-5$, $f(x, y)$ reaches its maximum value of 7.