Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form A x 2 + B x y + C y 2 + D x + E y + F = 0 A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0 A x 2 + B x y + C y 2 + D x + E y + F = 0 into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the x ′ {x}^{\prime } x ′ and y ′ {y}^{\prime } y ′ coordinate system without the x ′ y ′ {x}^{\prime }{y}^{\prime } x ′ y ′ term, by rotating the axes by a measure of θ \theta θ that satisfies
Solution
First, we find
cot ( 2 θ ) \cot \left(2\theta \right) cot ( 2 θ ) .
8 x 2 − 12 x y + 17 y 2 = 20 ⇒ A = 8 , B = − 12 and C = 17 cot ( 2 θ ) = A − C B = 8 − 17 − 12 cot ( 2 θ ) = − 9 − 12 = 3 4 \begin{array}{l}8{x}^{2}-12xy+17{y}^{2}=20\Rightarrow A=8,B=-12\text{and}C=17\hfill \\ \text{ }\cot \left(2\theta \right)=\frac{A-C}{B}=\frac{8 - 17}{-12}\hfill \\ \text{ }\cot \left(2\theta \right)=\frac{-9}{-12}=\frac{3}{4}\hfill \end{array} 8 x 2 − 12 x y + 17 y 2 = 20 ⇒ A = 8 , B = − 12 and C = 17 cot ( 2 θ ) = B A − C = − 12 8 − 17 cot ( 2 θ ) = − 12 − 9 = 4 3
Figure 7
cot ( 2 θ ) = 3 4 = adjacent opposite \cot \left(2\theta \right)=\frac{3}{4}=\frac{\text{adjacent}}{\text{opposite}} cot ( 2 θ ) = 4 3 = opposite adjacent
So the hypotenuse is
3 2 + 4 2 = h 2 9 + 16 = h 2 25 = h 2 h = 5 \begin{array}{r}\hfill {3}^{2}+{4}^{2}={h}^{2}\\ \hfill 9+16={h}^{2}\\ \hfill 25={h}^{2}\\ \hfill h=5\end{array} 3 2 + 4 2 = h 2 9 + 16 = h 2 25 = h 2 h = 5
Next, we find
sin θ \sin \text{ }\theta sin θ and
cos θ \cos \text{ }\theta cos θ .
sin θ = 1 − cos ( 2 θ ) 2 = 1 − 3 5 2 = 5 5 − 3 5 2 = 5 − 3 5 ⋅ 1 2 = 2 10 = 1 5 sin θ = 1 5 cos θ = 1 + cos ( 2 θ ) 2 = 1 + 3 5 2 = 5 5 + 3 5 2 = 5 + 3 5 ⋅ 1 2 = 8 10 = 4 5 cos θ = 2 5 \begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ \sin \text{ }\theta =\sqrt{\frac{1-\cos \left(2\theta \right)}{2}}=\sqrt{\frac{1-\frac{3}{5}}{2}}=\sqrt{\frac{\frac{5}{5}-\frac{3}{5}}{2}}=\sqrt{\frac{5 - 3}{5}\cdot \frac{1}{2}}=\sqrt{\frac{2}{10}}=\sqrt{\frac{1}{5}}\hfill \end{array}\hfill \\ \sin \text{ }\theta =\frac{1}{\sqrt{5}}\hfill \\ \cos \text{ }\theta =\sqrt{\frac{1+\cos \left(2\theta \right)}{2}}=\sqrt{\frac{1+\frac{3}{5}}{2}}=\sqrt{\frac{\frac{5}{5}+\frac{3}{5}}{2}}=\sqrt{\frac{5+3}{5}\cdot \frac{1}{2}}=\sqrt{\frac{8}{10}}=\sqrt{\frac{4}{5}}\hfill \\ \cos \text{ }\theta =\frac{2}{\sqrt{5}}\hfill \end{array} sin θ = 2 1 − c o s ( 2 θ ) = 2 1 − 5 3 = 2 5 5 − 5 3 = 5 5 − 3 ⋅ 2 1 = 10 2 = 5 1 sin θ = 5 1 cos θ = 2 1 + c o s ( 2 θ ) = 2 1 + 5 3 = 2 5 5 + 5 3 = 5 5 + 3 ⋅ 2 1 = 10 8 = 5 4 cos θ = 5 2
Substitute the values of
sin θ \sin \text{ }\theta sin θ and
cos θ \cos \text{ }\theta cos θ into
x = x ′ cos θ − y ′ sin θ x={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta x = x ′ cos θ − y ′ sin θ and
y = x ′ sin θ + y ′ cos θ y={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta y = x ′ sin θ + y ′ cos θ .
x = x ′ cos θ − y ′ sin θ x = x ′ ( 2 5 ) − y ′ ( 1 5 ) x = 2 x ′ − y ′ 5 \begin{array}{l}\hfill \\ \begin{array}{l}x={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta \hfill \\ x={x}^{\prime }\left(\frac{2}{\sqrt{5}}\right)-{y}^{\prime }\left(\frac{1}{\sqrt{5}}\right)\hfill \\ x=\frac{2{x}^{\prime }-{y}^{\prime }}{\sqrt{5}}\hfill \end{array}\hfill \end{array} x = x ′ cos θ − y ′ sin θ x = x ′ ( 5 2 ) − y ′ ( 5 1 ) x = 5 2 x ′ − y ′
and
y = x ′ sin θ + y ′ cos θ y = x ′ ( 1 5 ) + y ′ ( 2 5 ) y = x ′ + 2 y ′ 5 \begin{array}{l}\begin{array}{l}\hfill \\ y={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta \hfill \end{array}\hfill \\ y={x}^{\prime }\left(\frac{1}{\sqrt{5}}\right)+{y}^{\prime }\left(\frac{2}{\sqrt{5}}\right)\hfill \\ y=\frac{{x}^{\prime }+2{y}^{\prime }}{\sqrt{5}}\hfill \end{array} y = x ′ sin θ + y ′ cos θ y = x ′ ( 5 1 ) + y ′ ( 5 2 ) y = 5 x ′ + 2 y ′
Substitute the expressions for
x x x and
y y y into in the given equation, and then simplify.
8 ( 2 x ′ − y ′ 5 ) 2 − 12 ( 2 x ′ − y ′ 5 ) ( x ′ + 2 y ′ 5 ) + 17 ( x ′ + 2 y ′ 5 ) 2 = 20 8 ( ( 2 x ′ − y ′ ) ( 2 x ′ − y ′ ) 5 ) − 12 ( ( 2 x ′ − y ′ ) ( x ′ + 2 y ′ ) 5 ) + 17 ( ( x ′ + 2 y ′ ) ( x ′ + 2 y ′ ) 5 ) = 20 8 ( 4 x ′ 2 − 4 x ′ y ′ + y ′ 2 ) − 12 ( 2 x ′ 2 + 3 x ′ y ′ − 2 y ′ 2 ) + 17 ( x ′ 2 + 4 x ′ y ′ + 4 y ′ 2 ) = 100 32 x ′ 2 − 32 x ′ y ′ + 8 y ′ 2 − 24 x ′ 2 − 36 x ′ y ′ + 24 y ′ 2 + 17 x ′ 2 + 68 x ′ y ′ + 68 y ′ 2 = 100 25 x ′ 2 + 100 y ′ 2 = 100 25 100 x ′ 2 + 100 100 y ′ 2 = 100 100 \begin{array}{l}\text{ }8{\left(\frac{2{x}^{\prime }-{y}^{\prime }}{\sqrt{5}}\right)}^{2}-12\left(\frac{2{x}^{\prime }-{y}^{\prime }}{\sqrt{5}}\right)\left(\frac{{x}^{\prime }+2{y}^{\prime }}{\sqrt{5}}\right)+17{\left(\frac{{x}^{\prime }+2{y}^{\prime }}{\sqrt{5}}\right)}^{2}=20\text{ }\hfill \\ \text{ }8\left(\frac{\left(2{x}^{\prime }-{y}^{\prime }\right)\left(2{x}^{\prime }-{y}^{\prime }\right)}{5}\right)-12\left(\frac{\left(2{x}^{\prime }-{y}^{\prime }\right)\left({x}^{\prime }+2{y}^{\prime }\right)}{5}\right)+17\left(\frac{\left({x}^{\prime }+2{y}^{\prime }\right)\left({x}^{\prime }+2{y}^{\prime }\right)}{5}\right)=20\text{ }\hfill \\ \text{ }8\left(4{x}^{\prime }{}^{2}-4{x}^{\prime }{y}^{\prime }+{y}^{\prime }{}^{2}\right)-12\left(2{x}^{\prime }{}^{2}+3{x}^{\prime }{y}^{\prime }-2{y}^{\prime }{}^{2}\right)+17\left({x}^{\prime }{}^{2}+4{x}^{\prime }{y}^{\prime }+4{y}^{\prime }{}^{2}\right)=100\hfill \\ 32{x}^{\prime }{}^{2}-32{x}^{\prime }{y}^{\prime }+8{y}^{\prime }{}^{2}-24{x}^{\prime }{}^{2}-36{x}^{\prime }{y}^{\prime }+24{y}^{\prime }{}^{2}+17{x}^{\prime }{}^{2}+68{x}^{\prime }{y}^{\prime }+68{y}^{\prime }{}^{2}=100\hfill \\ \text{ }25{x}^{\prime }{}^{2}+100{y}^{\prime }{}^{2}=100\text{ }\hfill \\ \text{ }\frac{25}{100}{x}^{\prime }{}^{2}+\frac{100}{100}{y}^{\prime }{}^{2}=\frac{100}{100} \hfill \end{array} 8 ( 5 2 x ′ − y ′ ) 2 − 12 ( 5 2 x ′ − y ′ ) ( 5 x ′ + 2 y ′ ) + 17 ( 5 x ′ + 2 y ′ ) 2 = 20 8 ( 5 ( 2 x ′ − y ′ ) ( 2 x ′ − y ′ ) ) − 12 ( 5 ( 2 x ′ − y ′ ) ( x ′ + 2 y ′ ) ) + 17 ( 5 ( x ′ + 2 y ′ ) ( x ′ + 2 y ′ ) ) = 20 8 ( 4 x ′ 2 − 4 x ′ y ′ + y ′ 2 ) − 12 ( 2 x ′ 2 + 3 x ′ y ′ − 2 y ′ 2 ) + 17 ( x ′ 2 + 4 x ′ y ′ + 4 y ′ 2 ) = 100 32 x ′ 2 − 32 x ′ y ′ + 8 y ′ 2 − 24 x ′ 2 − 36 x ′ y ′ + 24 y ′ 2 + 17 x ′ 2 + 68 x ′ y ′ + 68 y ′ 2 = 100 25 x ′ 2 + 100 y ′ 2 = 100 100 25 x ′ 2 + 100 100 y ′ 2 = 100 100
Write the equations with
x ′ {x}^{\prime } x ′ and
y ′ {y}^{\prime } y ′ in the standard form with respect to the new coordinate system.
x ′ 2 4 + y ′ 2 1 = 1 \frac{{{x}^{\prime }}^{2}}{4}+\frac{{{y}^{\prime }}^{2}}{1}=1 4 x ′ 2 + 1 y ′ 2 = 1
Figure 8 shows the graph of the ellipse.
Figure 8
Example 4: Graphing an Equation That Has No x′y′ Terms
Graph the following equation relative to the
x ′ y ′ {x}^{\prime }{y}^{\prime } x ′ y ′ system:
x 2 + 12 x y − 4 y 2 = 30 {x}^{2}+12xy - 4{y}^{2}=30 x 2 + 12 x y − 4 y 2 = 30
Solution
First, we find
cot ( 2 θ ) \cot \left(2\theta \right) cot ( 2 θ ) .
x 2 + 12 x y − 4 y 2 = 20 ⇒ A = 1 , B = 12 , and C = − 4 {x}^{2}+12xy - 4{y}^{2}=20\Rightarrow A=1,\text{ }B=12,\text{and }C=-4 x 2 + 12 x y − 4 y 2 = 20 ⇒ A = 1 , B = 12 , and C = − 4
cot ( 2 θ ) = A − C B cot ( 2 θ ) = 1 − ( − 4 ) 12 cot ( 2 θ ) = 5 12 \begin{array}{l}\cot \left(2\theta \right)=\frac{A-C}{B}\hfill \\ \cot \left(2\theta \right)=\frac{1-\left(-4\right)}{12}\hfill \\ \cot \left(2\theta \right)=\frac{5}{12}\hfill \end{array} cot ( 2 θ ) = B A − C cot ( 2 θ ) = 12 1 − ( − 4 ) cot ( 2 θ ) = 12 5
Because
cot ( 2 θ ) = 5 12 \cot \left(2\theta \right)=\frac{5}{12} cot ( 2 θ ) = 12 5 , we can draw a reference triangle as in Figure 9.
Figure 9
cot ( 2 θ ) = 5 12 = adjacent opposite \cot \left(2\theta \right)=\frac{5}{12}=\frac{\text{adjacent}}{\text{opposite}} cot ( 2 θ ) = 12 5 = opposite adjacent
Thus, the hypotenuse is
5 2 + 12 2 = h 2 25 + 144 = h 2 169 = h 2 h = 13 \begin{array}{r}\hfill {5}^{2}+{12}^{2}={h}^{2}\\ \hfill 25+144={h}^{2}\\ \hfill 169={h}^{2}\\ \hfill h=13\end{array} 5 2 + 12 2 = h 2 25 + 144 = h 2 169 = h 2 h = 13
Next, we find
sin θ \sin \text{ }\theta sin θ and
cos θ \cos \text{ }\theta cos θ . We will use half-angle identities.
sin θ = 1 − cos ( 2 θ ) 2 = 1 − 5 13 2 = 13 13 − 5 13 2 = 8 13 ⋅ 1 2 = 2 13 cos θ = 1 + cos ( 2 θ ) 2 = 1 + 5 13 2 = 13 13 + 5 13 2 = 18 13 ⋅ 1 2 = 3 13 \begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ \sin \text{ }\theta =\sqrt{\frac{1-\cos \left(2\theta \right)}{2}}=\sqrt{\frac{1-\frac{5}{13}}{2}}=\sqrt{\frac{\frac{13}{13}-\frac{5}{13}}{2}}=\sqrt{\frac{8}{13}\cdot \frac{1}{2}}=\frac{2}{\sqrt{13}}\hfill \end{array}\hfill \\ \cos \text{ }\theta =\sqrt{\frac{1+\cos \left(2\theta \right)}{2}}=\sqrt{\frac{1+\frac{5}{13}}{2}}=\sqrt{\frac{\frac{13}{13}+\frac{5}{13}}{2}}=\sqrt{\frac{18}{13}\cdot \frac{1}{2}}=\frac{3}{\sqrt{13}}\hfill \end{array} sin θ = 2 1 − c o s ( 2 θ ) = 2 1 − 13 5 = 2 13 13 − 13 5 = 13 8 ⋅ 2 1 = 13 2 cos θ = 2 1 + c o s ( 2 θ ) = 2 1 + 13 5 = 2 13 13 + 13 5 = 13 18 ⋅ 2 1 = 13 3
Now we find
x x x and
y . y\text{.\hspace{0.17em}} y .
x = x ′ cos θ − y ′ sin θ x = x ′ ( 3 13 ) − y ′ ( 2 13 ) x = 3 x ′ − 2 y ′ 13 \begin{array}{l}\hfill \\ x={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta \hfill \\ x={x}^{\prime }\left(\frac{3}{\sqrt{13}}\right)-{y}^{\prime }\left(\frac{2}{\sqrt{13}}\right)\hfill \\ x=\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}}\hfill \end{array} x = x ′ cos θ − y ′ sin θ x = x ′ ( 13 3 ) − y ′ ( 13 2 ) x = 13 3 x ′ − 2 y ′
and
y = x ′ sin θ + y ′ cos θ y = x ′ ( 2 13 ) + y ′ ( 3 13 ) y = 2 x ′ + 3 y ′ 13 \begin{array}{l}\hfill \\ y={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta \hfill \\ y={x}^{\prime }\left(\frac{2}{\sqrt{13}}\right)+{y}^{\prime }\left(\frac{3}{\sqrt{13}}\right)\hfill \\ y=\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}}\hfill \end{array} y = x ′ sin θ + y ′ cos θ y = x ′ ( 13 2 ) + y ′ ( 13 3 ) y = 13 2 x ′ + 3 y ′
Now we substitute
x = 3 x ′ − 2 y ′ 13 x=\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}} x = 13 3 x ′ − 2 y ′ and
y = 2 x ′ + 3 y ′ 13 y=\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}} y = 13 2 x ′ + 3 y ′ into
x 2 + 12 x y − 4 y 2 = 30 {x}^{2}+12xy - 4{y}^{2}=30 x 2 + 12 x y − 4 y 2 = 30 .
( 3 x ′ − 2 y ′ 13 ) 2 + 12 ( 3 x ′ − 2 y ′ 13 ) ( 2 x ′ + 3 y ′ 13 ) − 4 ( 2 x ′ + 3 y ′ 13 ) 2 = 30 ( 1 13 ) [ ( 3 x ′ − 2 y ′ ) 2 + 12 ( 3 x ′ − 2 y ′ ) ( 2 x ′ + 3 y ′ ) − 4 ( 2 x ′ + 3 y ′ ) 2 ] = 30 Factor . ( 1 13 ) [ 9 x ′ 2 − 12 x ′ y ′ + 4 y ′ 2 + 12 ( 6 x ′ 2 + 5 x ′ y ′ − 6 y ′ 2 ) − 4 ( 4 x ′ 2 + 12 x ′ y ′ + 9 y ′ 2 ) ] = 30 Multiply . ( 1 13 ) [ 9 x ′ 2 − 12 x ′ y ′ + 4 y ′ 2 + 72 x ′ 2 + 60 x ′ y ′ − 72 y ′ 2 − 16 x ′ 2 − 48 x ′ y ′ − 36 y ′ 2 ] = 30 Distribute . ( 1 13 ) [ 65 x ′ 2 − 104 y ′ 2 ] = 30 Combine like terms . 65 x ′ 2 − 104 y ′ 2 = 390 Multiply . x ′ 2 6 − 4 y ′ 2 15 = 1 Divide by 390 . \begin{array}{llll}\text{ }{\left(\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}}\right)}^{2}+12\left(\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}}\right)\left(\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}}\right)-4{\left(\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}}\right)}^{2}=30\hfill & \hfill & \hfill & \hfill \\ \text{ }\left(\frac{1}{13}\right)\left[{\left(3{x}^{\prime }-2{y}^{\prime }\right)}^{2}+12\left(3{x}^{\prime }-2{y}^{\prime }\right)\left(2{x}^{\prime }+3{y}^{\prime }\right)-4{\left(2{x}^{\prime }+3{y}^{\prime }\right)}^{2}\right]=30 \hfill & \hfill & \hfill & \text{Factor}.\hfill \\ \left(\frac{1}{13}\right)\left[9{x}^{\prime }{}^{2}-12{x}^{\prime }{y}^{\prime }+4{y}^{\prime }{}^{2}+12\left(6{x}^{\prime }{}^{2}+5{x}^{\prime }{y}^{\prime }-6{y}^{\prime }{}^{2}\right)-4\left(4{x}^{\prime }{}^{2}+12{x}^{\prime }{y}^{\prime }+9{y}^{\prime }{}^{2}\right)\right]=30\hfill & \hfill & \hfill & \text{Multiply}.\hfill \\ \text{ }\left(\frac{1}{13}\right)\left[9{x}^{\prime }{}^{2}-12{x}^{\prime }{y}^{\prime }+4{y}^{\prime }{}^{2}+72{x}^{\prime }{}^{2}+60{x}^{\prime }{y}^{\prime }-72{y}^{\prime }{}^{2}-16{x}^{\prime }{}^{2}-48{x}^{\prime }{y}^{\prime }-36{y}^{\prime }{}^{2}\right]=30\hfill & \hfill & \hfill & \text{Distribute}.\hfill \\ \text{ }\text{ }\left(\frac{1}{13}\right)\left[65{x}^{\prime }{}^{2}-104{y}^{\prime }{}^{2}\right]=30\hfill & \hfill & \hfill & \text{Combine like terms}.\hfill \\ \text{ }65{x}^{\prime }{}^{2}-104{y}^{\prime }{}^{2}=390\hfill & \hfill & \hfill & \text{Multiply}.\text{ }\hfill \\ \text{ }\frac{{x}^{\prime }{}^{2}}{6}-\frac{4{y}^{\prime }{}^{2}}{15}=1 \hfill & \hfill & \hfill & \text{Divide by 390}.\hfill \end{array} ( 13 3 x ′ − 2 y ′ ) 2 + 12 ( 13 3 x ′ − 2 y ′ ) ( 13 2 x ′ + 3 y ′ ) − 4 ( 13 2 x ′ + 3 y ′ ) 2 = 30 ( 13 1 ) [ ( 3 x ′ − 2 y ′ ) 2 + 12 ( 3 x ′ − 2 y ′ ) ( 2 x ′ + 3 y ′ ) − 4 ( 2 x ′ + 3 y ′ ) 2 ] = 30 ( 13 1 ) [ 9 x ′ 2 − 12 x ′ y ′ + 4 y ′ 2 + 12 ( 6 x ′ 2 + 5 x ′ y ′ − 6 y ′ 2 ) − 4 ( 4 x ′ 2 + 12 x ′ y ′ + 9 y ′ 2 ) ] = 30 ( 13 1 ) [ 9 x ′ 2 − 12 x ′ y ′ + 4 y ′ 2 + 72 x ′ 2 + 60 x ′ y ′ − 72 y ′ 2 − 16 x ′ 2 − 48 x ′ y ′ − 36 y ′ 2 ] = 30 ( 13 1 ) [ 65 x ′ 2 − 104 y ′ 2 ] = 30 65 x ′ 2 − 104 y ′ 2 = 390 6 x ′ 2 − 15 4 y ′ 2 = 1 Factor . Multiply . Distribute . Combine like terms . Multiply . Divide by 390 .
Figure 10 shows the graph of the hyperbola
x ′ 2 6 − 4 y ′ 2 15 = 1. \frac{{{x}^{\prime }}^{2}}{6}-\frac{4{{y}^{\prime }}^{2}}{15}=1.\text{ } 6 x ′ 2 − 15 4 y ′ 2 = 1.
Figure 10