Practice!

Sample Problem

Sketch the graph of the rotated conic x2-4xy+y2+9=0.

Solution

Based on Ax2+Bxy+Cy2+Dx+Ey+F=0,

we know that x2-4xy+y2+9=0 has A=1, B=-4, and C=1 and therefore we have

cot 2θ=(A-C)/B=(1-1)/-4=0 → 2θ=π/2 → θ=π/4 (sin π/4=cos π/4=√2/2)

x=x’cosθ-y’sinθ=√2/2x’-√2/2y’

y=x’sinθ+y’cosθ=√2/2x’+√2/2y’

The equation in the x’y’ system is obtained by substituting these expressions in the equation

x2-4xy+y2+9=0

x2-4xy+y2+9=0

(√2/2x’-√2/2y’)2-4(√2/2x’-√2/2y’)(√2/2x’+√2/2y’)+(√2/2x’+√2/2y’)2=-9

1/2x’2-x’y’+1/2y’2-4(1/2x’2-1/2y’2)+(1/2x’2+x’y’+1/2y’2)=-9

1/2x’2-x’y’+1/2y’2-2x’2+2y’2+1/2x’2+x’y’+1/2y’2=-9

-x’2+3y’2=-9

x’2-3y’2=9

This is a rotated hyperbola with a horizontal transverse axis since the x2 term is positive. The rotated hyperbola is centered at (0, 0) and has vertices at (3, 0) and (-3, 0) [since a2=9 – a=3, which is the distance from center to vertices] and foci at (2√3, 0) and (-2√3, 0) [since c2=a2+b2=9+3=12 → c=√12=2√3, which is the distance from the center to foci]

Therefore, the graph of this hyperbola looks like