15. As shown in the figure, the curve $C$ is composed of the upper half of the ellipse $C_{1}: \frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1(a>b>0, y \geqslant 0)$ and part of the parabola $C_{2}: y=-x^{2}+1(y \leqslant 0)$, which are connected at points $A, B$. The eccentricity of $C_{1}$ is $\frac{\sqrt{3}}{2}$.
(1) Find the values of $a, b$;
(2) A line $l$ passing through point $B$ intersects $C_{1}, C_{2}$ at points $P, Q$ (both different from points $A, B$). If $A P \perp A Q$, find the equation of the line $l$.
(1) $b=1, e^{2}=1-\frac{b^{2}}{a^{2}}=\frac{3}{4} \Rightarrow a=2$.
(2) Let the line $l: y=k(x-1)$, from (1) we know the equation of the ellipse is $\frac{y^{2}}{4}+x^{2}=1$, combining
$$
\left\{\begin{array}{l}
y=k(x-1), \\
4 x^{2}+y^{2}=4
\end{array} \Rightarrow\left(k^{2}+4\right) x^{2}-2 k^{2} x+k^{2}-4=0\right. \text {. }
$$

Solving, we get $x_{P}=\frac{k^{2}-4}{k^{2}+4}, y_{P}=\frac{-8 k}{k^{2}+4} \Rightarrow P\left(\frac{k^{2}-4}{k^{2}+4}, \frac{-8 k}{k^{2}+4}\right)$.
Similarly, combining $\left\{\begin{array}{l}y=k(x-1), \\ y=-x^{2}+1\end{array} \Rightarrow x^{2}+k x-k-1=0 \Rightarrow Q\left(-k-1,-k^{2}-2 k\right)\right.$.
Thus, $\overrightarrow{A P} \cdot \overrightarrow{A Q}=\left(\frac{k^{2}-4}{k^{2}+4}+1, \frac{-8 k}{k^{2}+4}\right) \cdot\left(-k-1+1,-k^{2}-2 k\right)$
$$
=\frac{-2 k^{3}}{k^{2}+4}+\frac{8 k^{2}(k+2)}{k^{2}+4}=\frac{2 k^{2}(3 k+8)}{k^{2}+4}=0 \Rightarrow k=-\frac{8}{3} \text {. }
$$

Therefore, the equation of the line $l$ is $y=-\frac{8}{3}(x-1) \Rightarrow 8 x+3 y-8=0$.