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
Aug 29, 2016

Answer:

#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.