\quad \begin{array}{l}\mathrm{p}=\mathrm{x} \cos \theta-\mathrm{y} \sin \theta \\ \mathrm{q}=\mathrm{x} \sin \theta+\mathrm{y} \cos \theta \\ \text { Given, } \mathrm{p}^{2}+4 \mathrm{pq}+\mathrm{q}^{2}=\mathrm{Ax}^{2}+\mathrm{By}^{2}\end{array} \end{aligned}$
Let us take $\theta=\frac{\pi}{4}$.
$p=x \cos \frac{\pi}{4}-y \cos \frac{\pi}{4}=\frac{x-y}{\sqrt{2}}$
$$
\begin{array}{l}
q=x \sin \frac{\pi}{4}+y \cos \frac{\pi}{4}=\frac{x+y}{\sqrt{2}} \\
p q=\frac{x^{2}-y^{2}}{2} \Rightarrow 2 p q=x^{2}-y^{2} \\
\Rightarrow 4 \mathrm{pq}=2 \mathrm{x}^{2}-2 \mathrm{y}^{2} \\
\text { Now, } \mathrm{p}^{2}+\mathrm{q}^{2}=\mathrm{x}^{2} \cos ^{2} \theta+\mathrm{y}^{2} \sin ^{2} \theta-2 \mathrm{xy} \cos \theta \sin \theta \\
+\mathrm{x}^{2} \sin ^{2} \theta+\mathrm{y}^{2} \cos ^{2} \theta+2 \mathrm{x} \mathrm{sin} \theta \cos \theta=\mathrm{x}^{2}+\mathrm{y}^{2} \\
\text { From }(1),(2), \mathrm{p}^{2}+\mathrm{q}^{2}+4 \mathrm{pq}=\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}^{2}-2 \mathrm{y}^{2} \\
=3 \mathrm{x}^{2}-\mathrm{y}^{2}
\end{array}
$$
Comparing this with the given form, we get
$$
\theta=\frac{\pi}{4}, A=3, B=-1
$$