Answer to Question #160250 in Linear Algebra for Nikhil Singh

Question

Reduce the quadratic form 5x^2-4xy+8y^2 to its orthogonal canonical form, clearly giving the transformations being used. Also give a rough sketch of the curve representing this canonical form.

Answer 1

$5x^2 -4xy +8y^2 = 1$ (instead of one, any number can be here, you did not specify it)

$(nx \pm my)^2 = c$ - canonical form

first of all we should find the square roots of the first and last terms:

$\sqrt{5x^2} = \sqrt{5}*x$

$\sqrt{8y^2} = 2\sqrt{2}*y$

-4 - should be multiplication of $\sqrt{5}$ and $2\sqrt{2}$ and + free term:

$\sqrt{5}*2\sqrt{2} = 2\sqrt{10}$

Answer: $(\sqrt{5}x + 2\sqrt{2}y)^2 - (4+2\sqrt{10})xy = 1$