The given equation is 4 x y − 3 y 2 = 64 4xy - 3y^2 = 64 4 x y − 3 y 2 = 64 . To eliminate the cross-product term x y xy x y , we rotate the coordinate axes by an angle θ \theta θ . The rotation formulas are
x = x ′ cos θ − y ′ sin θ x = x' \cos \theta - y' \sin \theta x = x ′ cos θ − y ′ sin θ y = x ′ sin θ + y ′ cos θ y = x' \sin \theta + y' \cos \theta y = x ′ sin θ + y ′ cos θ
Substituting these into the equation gives
4 ( x ′ cos θ − y ′ sin θ ) ( x ′ sin θ + y ′ cos θ ) − 3 ( x ′ sin θ + y ′ cos θ ) 2 = 64 4(x' \cos \theta - y' \sin \theta)(x' \sin \theta + y' \cos \theta) - 3(x' \sin \theta + y' \cos \theta)^2 = 64 4 ( x ′ cos θ − y ′ sin θ ) ( x ′ sin θ + y ′ cos θ ) − 3 ( x ′ sin θ + y ′ cos θ ) 2 = 64 4 ( x ′ 2 sin θ cos θ + x ′ y ′ cos 2 θ − x ′ y ′ sin 2 θ − y ′ 2 sin θ cos θ ) − 3 ( x ′ 2 sin 2 θ + 2 x ′ y ′ sin θ cos θ + y ′ 2 cos 2 θ ) = 64 4(x'^2 \sin \theta \cos \theta + x'y' \cos^2 \theta - x'y' \sin^2 \theta - y'^2 \sin \theta \cos \theta) - 3(x'^2 \sin^2 \theta + 2x'y' \sin \theta \cos \theta + y'^2 \cos^2 \theta) = 64 4 ( x ′2 sin θ cos θ + x ′ y ′ cos 2 θ − x ′ y ′ sin 2 θ − y ′2 sin θ cos θ ) − 3 ( x ′2 sin 2 θ + 2 x ′ y ′ sin θ cos θ + y ′2 cos 2 θ ) = 64 4 x ′ 2 sin θ cos θ + 4 x ′ y ′ ( cos 2 θ − sin 2 θ ) − 4 y ′ 2 sin θ cos θ − 3 x ′ 2 sin 2 θ − 6 x ′ y ′ sin θ cos θ − 3 y ′ 2 cos 2 θ = 64 4x'^2 \sin \theta \cos \theta + 4x'y' (\cos^2 \theta - \sin^2 \theta) - 4y'^2 \sin \theta \cos \theta - 3x'^2 \sin^2 \theta - 6x'y' \sin \theta \cos \theta - 3y'^2 \cos^2 \theta = 64 4 x ′2 sin θ cos θ + 4 x ′ y ′ ( cos 2 θ − sin 2 θ ) − 4 y ′2 sin θ cos θ − 3 x ′2 sin 2 θ − 6 x ′ y ′ sin θ cos θ − 3 y ′2 cos 2 θ = 64 x ′ 2 ( 4 sin θ cos θ − 3 sin 2 θ ) + x ′ y ′ ( 4 cos 2 θ − 4 sin 2 θ − 6 sin θ cos θ ) + y ′ 2 ( − 4 sin θ cos θ − 3 cos 2 θ ) = 64 x'^2(4 \sin \theta \cos \theta - 3 \sin^2 \theta) + x'y'(4 \cos^2 \theta - 4 \sin^2 \theta - 6 \sin \theta \cos \theta) + y'^2(-4 \sin \theta \cos \theta - 3 \cos^2 \theta) = 64 x ′2 ( 4 sin θ cos θ − 3 sin 2 θ ) + x ′ y ′ ( 4 cos 2 θ − 4 sin 2 θ − 6 sin θ cos θ ) + y ′2 ( − 4 sin θ cos θ − 3 cos 2 θ ) = 64
We want the coefficient of the x ′ y ′ x'y' x ′ y ′ term to be zero. 4 cos 2 θ − 4 sin 2 θ − 6 sin θ cos θ = 0 4 \cos^2 \theta - 4 \sin^2 \theta - 6 \sin \theta \cos \theta = 0 4 cos 2 θ − 4 sin 2 θ − 6 sin θ cos θ = 0 4 ( cos 2 θ − sin 2 θ ) − 3 ( 2 sin θ cos θ ) = 0 4 (\cos^2 \theta - \sin^2 \theta) - 3 (2 \sin \theta \cos \theta) = 0 4 ( cos 2 θ − sin 2 θ ) − 3 ( 2 sin θ cos θ ) = 0 4 cos ( 2 θ ) − 3 sin ( 2 θ ) = 0 4 \cos(2\theta) - 3 \sin(2\theta) = 0 4 cos ( 2 θ ) − 3 sin ( 2 θ ) = 0 4 cos ( 2 θ ) = 3 sin ( 2 θ ) 4 \cos(2\theta) = 3 \sin(2\theta) 4 cos ( 2 θ ) = 3 sin ( 2 θ ) tan ( 2 θ ) = 4 3 \tan(2\theta) = \frac{4}{3} tan ( 2 θ ) = 3 4
Let 2 θ = α 2\theta = \alpha 2 θ = α , so tan ( α ) = 4 3 \tan(\alpha) = \frac{4}{3} tan ( α ) = 3 4 . Then sin α = 4 5 \sin \alpha = \frac{4}{5} sin α = 5 4 and cos α = 3 5 \cos \alpha = \frac{3}{5} cos α = 5 3 . We have the identities:
sin 2 θ = 1 − cos ( 2 θ ) 2 \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} sin 2 θ = 2 1 − c o s ( 2 θ ) cos 2 θ = 1 + cos ( 2 θ ) 2 \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} cos 2 θ = 2 1 + c o s ( 2 θ ) sin θ cos θ = sin ( 2 θ ) 2 \sin \theta \cos \theta = \frac{\sin(2\theta)}{2} sin θ cos θ = 2 s i n ( 2 θ ) cos ( 2 θ ) = 3 5 \cos(2\theta) = \frac{3}{5} cos ( 2 θ ) = 5 3 sin ( 2 θ ) = 4 5 \sin(2\theta) = \frac{4}{5} sin ( 2 θ ) = 5 4
sin 2 θ = 1 − 3 5 2 = 2 5 2 = 1 5 \sin^2 \theta = \frac{1 - \frac{3}{5}}{2} = \frac{\frac{2}{5}}{2} = \frac{1}{5} sin 2 θ = 2 1 − 5 3 = 2 5 2 = 5 1 cos 2 θ = 1 + 3 5 2 = 8 5 2 = 4 5 \cos^2 \theta = \frac{1 + \frac{3}{5}}{2} = \frac{\frac{8}{5}}{2} = \frac{4}{5} cos 2 θ = 2 1 + 5 3 = 2 5 8 = 5 4 sin θ = 1 5 \sin \theta = \frac{1}{\sqrt{5}} sin θ = 5 1 cos θ = 2 5 \cos \theta = \frac{2}{\sqrt{5}} cos θ = 5 2 sin θ cos θ = 1 5 ⋅ 2 5 = 2 5 \sin \theta \cos \theta = \frac{1}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}} = \frac{2}{5} sin θ cos θ = 5 1 ⋅ 5 2 = 5 2
The equation becomes
x ′ 2 ( 4 sin θ cos θ − 3 sin 2 θ ) + y ′ 2 ( − 4 sin θ cos θ − 3 cos 2 θ ) = 64 x'^2(4 \sin \theta \cos \theta - 3 \sin^2 \theta) + y'^2(-4 \sin \theta \cos \theta - 3 \cos^2 \theta) = 64 x ′2 ( 4 sin θ cos θ − 3 sin 2 θ ) + y ′2 ( − 4 sin θ cos θ − 3 cos 2 θ ) = 64 x ′ 2 ( 4 ( 2 5 ) − 3 ( 1 5 ) ) + y ′ 2 ( − 4 ( 2 5 ) − 3 ( 4 5 ) ) = 64 x'^2(4(\frac{2}{5}) - 3(\frac{1}{5})) + y'^2(-4(\frac{2}{5}) - 3(\frac{4}{5})) = 64 x ′2 ( 4 ( 5 2 ) − 3 ( 5 1 )) + y ′2 ( − 4 ( 5 2 ) − 3 ( 5 4 )) = 64 x ′ 2 ( 8 5 − 3 5 ) + y ′ 2 ( − 8 5 − 12 5 ) = 64 x'^2(\frac{8}{5} - \frac{3}{5}) + y'^2(-\frac{8}{5} - \frac{12}{5}) = 64 x ′2 ( 5 8 − 5 3 ) + y ′2 ( − 5 8 − 5 12 ) = 64 x ′ 2 ( 5 5 ) + y ′ 2 ( − 20 5 ) = 64 x'^2(\frac{5}{5}) + y'^2(-\frac{20}{5}) = 64 x ′2 ( 5 5 ) + y ′2 ( − 5 20 ) = 64 x ′ 2 − 4 y ′ 2 = 64 x'^2 - 4y'^2 = 64 x ′2 − 4 y ′2 = 64 x ′ 2 64 − y ′ 2 16 = 1 \frac{x'^2}{64} - \frac{y'^2}{16} = 1 64 x ′2 − 16 y ′2 = 1
This is a hyperbola with a 2 = 64 a^2 = 64 a 2 = 64 and b 2 = 16 b^2 = 16 b 2 = 16 . a = 8 a = 8 a = 8 and b = 4 b = 4 b = 4 . The vertices are ( ± 8 , 0 ) (\pm 8, 0) ( ± 8 , 0 ) in the x ′ y ′ x'y' x ′ y ′ coordinate system. The asymptotes are y ′ = ± b a x ′ = ± 4 8 x ′ = ± 1 2 x ′ y' = \pm \frac{b}{a} x' = \pm \frac{4}{8} x' = \pm \frac{1}{2} x' y ′ = ± a b x ′ = ± 8 4 x ′ = ± 2 1 x ′ . The angle of rotation satisfies tan ( 2 θ ) = 4 3 \tan(2\theta) = \frac{4}{3} tan ( 2 θ ) = 3 4 . So 2 θ = arctan ( 4 3 ) ≈ 53.13 ∘ 2\theta = \arctan(\frac{4}{3}) \approx 53.13^{\circ} 2 θ = arctan ( 3 4 ) ≈ 53.1 3 ∘ , and θ ≈ 26.565 ∘ \theta \approx 26.565^{\circ} θ ≈ 26.56 5 ∘ . 