Parallel and perpendicular

The symmetric matrix of the quadratic form $q(x,y)=x^2-2xy+3y^2$ is $A=\begin{bmatrix}1&-1\\-1&3\end{bmatrix}$ with the eigenvalues $a=2+\sqrt{2}$ and $b=2-\sqrt{2}$ . Now $\frac{a}{b}=3+2\sqrt{2}\approx 5.828$ . The answer we seek is $\boxed{5828}$

Solve the simultaneous equations, $x^2-2xy+3y^2=k(x^2+y^2)$ which we can rewrite it as a quadratic in $x$ : $(1-k)x^2-2xy+(3-k)y^2=0$ . As the two curves touch each other, the discriminant is zero. Hence $4y^2-4(1-k)(3-k)y^2=0$ which means that $k^2-4k+2=0$ and hence $k=2\pm\sqrt{2}$ . Thus $\frac{a}{b}=\frac{2+\sqrt{2}}{2-\sqrt{2}}\approx 5.828$ . So $\left \lfloor{\dfrac{1000a}{b}}\right\rfloor=5828$

Any point lying on the circle would be of form * $(cos(x),sin(x))$ * so just satisfy the point in equation 2 we get

$sin(2x)+cos(2x)=2-k$ we want now it to be satisfied for only 2 points. therefore applying boundary line condition

$2-k=sqrt(2)$

$2-k=-sqrt(2)$

$k=2-sqrt(2)$

$k=2+sqrt(2)$

Like Dr Bretscher's, but using a symbolic algebra system. He reminded me of theory that I hadn't seen since the 70's when I studied principal components, just a day or two ago.