Finding real affine change of coordinates efficiently Consider the two equations
$$13x^2 - 10xy + 13y^2 = 1$$
$$4u^2 + 9v^2 = 1$$ 
What is a better way to find the change of coordinates than setting $x = au + bv + e$ and $y = du+bv+f$ and doing some terrible computations? 
When I studied multivariable calculus, I did just that, but I suppose there should be a better way. This question is from Algebraic Geometry by Garrity. 
 A: Hint: think geometrically. What are these figures? What does that suggest? (Consequences under the spoiler)

 They're both ellipses. So we can rotate the axes of one to lie on the axes of the other, and then scale in those directions, and that will give us our coordinate transformation. 

A full solution:

 The equation in $u$ and $v$ is an ellipse with major/minor axes of lengths $1/2$ and $1/3$ on the $u$ and $v$ axes, respectively. The equation in $x$ and $y$ is an ellipse with major/minor axes of length $1/6$ and $1/4$ in the $(1,-1)$ and $(1,1)$ directions, respectively. So we need to rotate by $\pi/4$ and dilate by a factor of $1/2$, so we write $x=(u+v)/2$ and $y=(u-v)/2$.

In general:

 Given two conics (the zero locus of one quadratic equation), we can use linear algebra to analyze whether we can get from one to the other. The key idea here is "diagonalizing the quadratic form": any quadratic equation over a field of characteristic not two can be represented by a matrix, where we take the entries on the diagonal to be the coefficients of the $x_i^2$ terms and the entries off the diagonal to be half the coefficients of the $x_ix_j$ term (and we add an extra row/column representing the constants). This matrix will be symmetric, and so it's diagonalizable over $\Bbb R$ - so after a change of basis our quadratic form will be $\sum \lambda_ix_i^2=c$, and it's clear how to manipulate these representations.

A: Note that the ellipse is centered at origin. The equations of its major and minor axes can be determined with $f(x,y) =13x^2 - 10xy + 13y^2$
$$\frac{f_y’}{f_y’}=\frac{-10x+26y}{26x-10y}= \frac yx\implies (x-y)(x+y)=0$$
i.e. they are at 45 degrees with respect with the $xy$-axes, thus the change of coordinates 
$$x=\frac c{\sqrt2}(x+y),\>\>\>\>\>y=\frac c{\sqrt2}(x-y)$$
with the scaling factor $c$. Substitute them into $13x^2 - 10xy + 13y^2=0$ and compare with $4u^2+9v^2=1$ to obtain $c= \frac1{\sqrt2}$.
