How can I find a linear transformation $\mathcal{T}$ that maps an ellipse to a circle?

Question

Let $\mathcal{T}(x,y)=(x',y')$ be a linear transformation.

The image of the ellipse $4x^2-4\sqrt{2}xy+11y^2=12$ under the transformation $\mathcal{T}$ is the circle $x'^2+y'^2=4$.

Answer 1

There is a standard way to do this kind of exercise; you have to "complete the squares": \begin{gather} 4x^2-4\sqrt 2 xy+ 11y^2 = 12\\ \left(2x-\sqrt 2 y\right)^2 -2y^2 + 11y^2 = 12\\ \left(2x-\sqrt 2 y\right)^2 + \left(3y\right)^2 = 12\\ \left(\frac{2x-\sqrt 2 y}{\sqrt 3}\right)^2 + \left(\frac{3y}{\sqrt 3}\right)^2 = 4\\ \end{gather}

So if a point $P=(\bar x,\bar y)$ lies in the ellipse $4x^2-4\sqrt 2 xy +11y^2 = 12$, then the point $Q=\left(\frac{2\bar x-\sqrt 2 \bar y}{\sqrt 3},\frac{3 \bar y}{\sqrt 3}\right)$ lies in the ellipse $x^2+y^2 = 4$.

Thanks to the above remark, the linear map you are looking for is: \begin{equation} \mathcal T (x,y) = \left(\frac{2 x-\sqrt 2 y}{\sqrt 3},\frac{3 y}{\sqrt 3}\right) \end{equation}