2. If the two foci of the hyperbola $L$ are exactly the two vertices of the ellipse $T: \frac{x^{2}}{16}+\frac{y^{2}}{9}=1$, and the two vertices of the hyperbola $L$ are exactly the two foci of the ellipse $T$, then the equation of the hyperbola $L$ is $\qquad$
Given $\frac{x^{2}}{7}-\frac{y^{2}}{9}=1$.
Analysis According to the conditions, the center of the hyperbola $L$ is at the origin, the real axis of symmetry is the $X$ axis. Suppose its equation is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, then its vertices are $( \pm a, 0)$, foci are $( \pm c, 0)$. The vertices of the major axis of the ellipse are $( \pm 4,0)$, and the foci are $( \pm \sqrt{7}, 0)$. Therefore, $a=\sqrt{7}, c=4$, so $b=\sqrt{c^{2}-a^{2}}=3$. Hence, the equation of the required hyperbola is $\frac{x^{2}}{7}-\frac{y^{2}}{9}=1$.