How do you graph the conic 13x^2+6sqrt3xy+7y^2-16=0 by first rotations the axis and eliminating the xy term?

1 Answer
Cesareo R.
Aug 29, 2016

16X^2+4Y^2-16=0 An ellipse.

Explanation:

13x^2+6sqrt3xy+7y^2-16=0 is equivalent to

((x),(y))^T((13,3sqrt(3)),(3sqrt(3),7))((x),(y)) - 16=0

Introducing a change of variables

((X),(Y)) = ((costheta, -Sintheta),(Sintheta, Costheta))((x),(y))

we get at

((X),(Y))^T((10 + 3 Cos2theta - 3 sqrt[3] Sin2theta, 3 (sqrt[3] Cos2theta + Sin2theta)),(3 (sqrt[3] Cos2theta + Sin2theta), 10 - 3 Cos2theta + 3 sqrt[3] Sin2theta))((X),(Y))-16=0

Choosing theta such that

3 (sqrt[3] Cos2theta + Sin2theta)=0

we have theta = pi/3 or theta = -pi/6

and the quadratic in this new set of coordinates reads

16X^2+4Y^2-16=0

or

4X^2+16Y^2-16=0

In both cases the quadratic is characterized as an ellipse.

Note. The quadratic kind can be determined computing the characteristic polynomial roots. In this case

M=((13,3sqrt(3)),(3sqrt(3),7))

has the characteristic polynomial

lambda^2-"trace"(M)lambda + det(M)=0

or

lambda^2-20lambda+64=0 with roots

lambda = {4, 16} characteririzing an ellipse.

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