Maximum and Minimum value of Quadratic form

Question

Consider the quadratic form $Q(\vec{x})=-2x^2+2\sqrt{33}xy+6y^2$ where $\vec{x}=\begin{bmatrix}x \\ y \end{bmatrix}$

It is easy to figure out the maximum and minimum value of $Q(\vec{x})$ when $\left \| \vec{x} \right \|=1$

I know how to do this, but what if $\left \| \vec{x} \right \|=5$???? then I don't know how to do it,

For example the minimum value of $Q(\vec{x})$ is $-5$ when $\left \| \vec{x} \right \|=1$, but what is the answer if $\left \| \vec{x} \right \|=5?$ Note: the correct answer is $-125$, but I have no idea why is it -125, any suggestion how to get this?

Answer 1

J.W. Tanner is correct

Suppose $\vec{x}=5\vec{x}_{0}$ where $||\vec{x}_{0}||=1$. Then

$$ \begin{align} Q(\vec{x})&=-2x^{2}+2\sqrt{33}xy+6y^{2}\\ &=-2(5x_{0})^{2}+2\sqrt{33}(5x_{0})(5y_{0})+6(y_{0})^{2}\\ &=25\left(-2x_{0}^{2}+2\sqrt{33}x_{0}y_{0}+6y_{0}^{2}\right)\\ &=25Q(\vec{x}_{0})\\ \\ \min{\left(Q(\vec{x})\right)}&=25\min{\left(Q(\vec{x}_{0})\right)} \end{align} $$