Sketch the graph of the rotated conic x^{2}-4xy+y^{2}+9=0.

Based on Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0,

we know that x^{2}-4xy+y^{2}+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

x^{2}-4xy+y^{2}+9=0

x^{2}-4xy+y^{2}+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 x^{2} term is positive. The rotated hyperbola is
centered at (0, 0) and has vertices at (3, 0) and (-3, 0) [since a^{2}=9
– a=3, which is the distance from center to vertices] and foci at (2√3, 0) and (-2√3, 0) [since c^{2}=a^{2}+b^{2}=9+3=12
→ c=√12=2√3, which is the distance from the
center to foci]

Therefore, the graph of this hyperbola looks like