Sketch the graph of the rotated conic x2-4xy+y2+9=0.
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)
The equation in the xy system is obtained by substituting these expressions in the equation
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