7. Let $A, B$ be the points of intersection of the hyperbola $W: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with the real axis, and $P(0,1)$ be a point outside the hyperbola. The lines $P A, P B$ intersect the hyperbola at another point $C, D$, respectively. The tangents at $C, D$ intersect at $Q$. If $\triangle Q C D$ is an equilateral triangle with an area of $\frac{16 \sqrt{3}}{27}$, then the equation of the hyperbola $W$ is $\qquad$.
$$
S_{\triangle Q C D}=\frac{\sqrt{3}}{4}|C D|^{2}=\frac{16 \sqrt{3}}{27} \Rightarrow|C D|=\frac{8 \sqrt{3}}{9} \Rightarrow x_{D}=\frac{4 \sqrt{3}}{9} \text {, }
$$

Also, $P B: y=-\frac{1}{a} x+1 \Rightarrow y_{D}=1-\frac{4 \sqrt{3}}{9 a}$, the equation of the tangent line to the hyperbola at point $\left(x_{0}, y_{0}\right)$ is $\frac{x_{0} x}{a^{2}}-\frac{y_{0} y}{b^{2}}=1$
$$
\Rightarrow k_{D Q}=\frac{b^{2} x_{D}}{a^{2} y_{D}}=\frac{\frac{4 \sqrt{3}}{9} b^{2}}{a^{2}-\frac{4 \sqrt{3}}{9} a}=-\sqrt{3} \Rightarrow a^{2}-\frac{4 \sqrt{3}}{9} a+\frac{4}{9} b^{2}=0 \text {. }
$$

Substituting point $D$ into the hyperbola equation gives $\frac{16}{27} b^{2}-\left(a-\frac{4 \sqrt{3}}{9}\right)^{2}=a^{2} b^{2}$,
then we have $-\frac{9}{4} \cdot \frac{16}{27}\left(a^{2}-\frac{4 \sqrt{3}}{9} a\right)-\left(a-\frac{4 \sqrt{3}}{9}\right)^{2}=-\frac{9}{4} \cdot a^{2}\left(a^{2}-\frac{4 \sqrt{3}}{9} a\right)$
$$
\begin{array}{l}
\Rightarrow \frac{4}{3} a+a-\frac{4 \sqrt{3}}{9}=\frac{9}{4} a^{3} \Rightarrow a^{3}-\frac{28}{27} a+\frac{16 \sqrt{3}}{81}=0 \\
\Rightarrow a^{3}-\frac{4 \sqrt{3}}{9} a^{2}+\frac{4 \sqrt{3}}{9} a^{2}-\frac{16}{27} a-\frac{4}{9} a+\frac{16 \sqrt{3}}{81}=\left(a-\frac{4 \sqrt{3}}{9}\right)\left(a^{2}+\frac{4 \sqrt{3}}{9} a-\frac{4}{9}\right)=0 \\
\Rightarrow\left(a-\frac{4 \sqrt{3}}{9}\right)\left(a-\frac{2 \sqrt{3}}{9}\right)\left(a+\frac{2 \sqrt{3}}{3}\right)=0 \Rightarrow a=\frac{2 \sqrt{3}}{9} .
\end{array}
$$

Thus, $b^{2}=-\frac{9}{4}\left(a^{2}-\frac{4 \sqrt{3}}{9} a\right)=\frac{1}{3}$, so the equation of the hyperbola $W$ is $\frac{27}{4} x^{2}-3 y^{2}=1$.