How do you graph the conic 2x^2-3xy-2y^2+10=0 by first rotations the axis and eliminating the xy term?

1 Answer
Cesareo R.
Sep 15, 2016

X^2-Y^2+4=0 which is a hyperbola.

Explanation:

With the rotation

R_(theta)=((ctheta,-stheta),(stheta,ctheta)), calling p = ((x),(y)) and P=((X),(Y)) we have first

p^T M p + 10 = 0 representing the conic, with M = ((2,-3/2),(-3/2,-2)). Here .^T represents transposition. Now introducing a change of coordinates

R_(theta)P = p we will have

P^TR_(theta)^TMR_(theta)P + 10 = 0

Calling M_(theta) = R_(theta)^TMR_(theta)=((2 Cos(2 theta) - 3 Cos(theta) Sin(theta), -3/2Cos(2 theta) - 2 Sin(2 theta)),(-3/2 Cos(2 theta) - 2 Sin(2 theta), -2 Cos(2 theta) + 3 Cos(theta) Sin(theta)))

Now, choosing theta such that -3/2 Cos(2 theta) - 2 Sin(2 theta)=0 we obtain

tan(2theta)=-3/4 and

theta = 1/2(-0.643501pmkpi) Choosing for k=0 we have

theta =-0.321751 and

M_(theta) = ((5/2,0),(0,-5/2)) and in the new coordinates, the conic reads

5/2X^2-5/2Y^2+10=0 or

X^2-Y^2+4=0 which is a hyperbola.

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