Values of quadratic form on unit circle

Question

We have the quadratic form $q(\begin{pmatrix}x\\y\end{pmatrix})=11x^2-16xy-y^2$.

Which values does $q$ take on the unit circle $x^2+y^2=1$?

I know that $q(x,y)$ is given by $q(x,y)=(x,y)\begin{pmatrix}11&-8\\-8&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$ and that $\begin{pmatrix}11&-8\\-8&-1\end{pmatrix}= \begin{pmatrix}-2&-1\\1&-2\end{pmatrix} \begin{pmatrix}15&0\\0&-5\end{pmatrix} \begin{pmatrix}-2&-1\\1&-2\end{pmatrix}^T$. But I don't know how I could use this further.

Answer 1

From Joseph Curwen's hint: $$\begin{align}q &= 11\cos^2\theta-16\cos \theta\sin \theta-\sin ^2\theta\\ &=12\cos^2\theta-16\cos \theta\sin \theta-1\end{align}$$ Now $$\sin 2\theta = 2\cos\theta\sin\theta\\\cos 2\theta = 2\cos^2 \theta - 1$$ So, $$\begin{align}q &= 6\cos 2\theta - 8\sin 2\theta + 5\\&=6\cos t - 8\sin t + 5\end{align}$$ Where $t = 2\theta$ (since $\theta$ was arbitrary, there is no reason to stick with it). Now set $q' = 0$ and find the maximum and minimum values of $q$. Since $q$ is continuous, it will take on every value between them.