7-05 Rotated Conics

Write a Rotated Conic in Standard Form

Classify \(xy = \frac{1}{2}\) and then write it in standard form.

Solution

Compare \(xy = \frac{1}{2}\) to Ax² + Bxy + Cy² + Dx + Ey + F = 0 and find that A = 0, B = 1, and C = 0.

Classify the conic by using the discriminant.

B² − 4AC

1² – 4(0)(0) = 1

Since B² − 4AC > 0, it is a hyperbola.

Write it in standard form by first finding the angle of rotation.

$$\cot ⁡2θ = \frac{A - C}{B} = \frac{0}{1} = 0$$

$$2θ = \frac{π}{2}$$

$$θ = \frac{π}{4}$$

Find the substitutions for x and y.

$$\begin{alignat}{2}x &= x′ \cos \frac{π}{4} &- y′ \sin \frac{π}{4}\\ y &= x′ \sin \frac{π}{4} &+ y′ \cos \frac {π}{4}\end{alignat}$$

$$\begin{alignat}{2}x &= \frac{\sqrt{2}}{2} x′ &- \frac{\sqrt{2}}{2} y′\\ y &= \frac{\sqrt{2}}{2} x′ &+ \frac{\sqrt{2}}{2} y′\end{alignat}$$

Substitute these into the original equation and simplify.

$$xy = \frac{1}{2}$$

$$\left(\frac{\sqrt{2}}{2} x′ - \frac{\sqrt{2}}{2} y′\right)\left(\frac{\sqrt{2}}{2} x′ + \frac{\sqrt{2}}{2} y′\right) = \frac{1}{2}$$

$$\frac{1}{2} (x′)^2 - \frac{1}{2} (x′y′) + \frac{1}{2} (x′y′) - \frac{1}{2} (y′)^2 = \frac{1}{2}$$

$$(x′)^2 - (y′)^2 = 1$$

From the equation, this conic is a hyperbola.

Graph a Rotated Conic

Classify \(3x^2 + \sqrt{3}xy + 2y^2 - 6 = 0\) and then sketch the graph.

Solution

Compare \(3x^2 + \sqrt{3}xy + 2y^2 - 6 = 0\) to Ax² + Bxy + Cy² + Dx + Ey + F = 0 and find that A = 3, B = \(\sqrt{3}\), and C = 2.

Classify the conic by using the discriminant.

B² − 4AC

$$\left(\sqrt{3}\right)^2 - 4(3)(2) = -21$$

Since B² − 4AC < 0, it is a ellipse.

Write it in standard form by first finding the angle of rotation.

$$\cot ⁡2θ = \frac{A - C}{B} = \frac{3 - 2}{\sqrt{3}} = \frac{1}{\sqrt{3}}$$

$$2θ = \frac{π}{3}$$

$$θ = \frac{π}{6}$$

Find the substitutions for x and y.

$$\begin{alignat}{2}x &= x′ \cos \frac{π}{6} &&- y′ \sin \frac{π}{6}\\ y &= x′ \sin \frac{π}{6} &&+ y′ \cos \frac{π}{6}\end{alignat}$$

$$\begin{alignat}{2}x &= \frac{\sqrt{3}}{2} x′ &&- \frac{1}{2} y′\\ y &= \frac{1}{2} x′ &&+ \frac{\sqrt{3}}{2} y′\end{alignat}$$

Substitute these into the original equation and simplify.

$$3\left(\frac{\sqrt{3}}{2} x′ - \frac{1}{2} y′\right)^2 + \sqrt{3} \left(\frac{\sqrt{3}}{2} x′ - \frac{1}{2} y′\right)\left(\frac{1}{2} x′ + \frac{\sqrt{3}}{2} y′\right) + 2\left(\frac{1}{2} x′ + \frac{\sqrt{3}}{2} y′\right)^2 - 6 = 0$$

$$3\left(\frac{3}{4} (x′)^2 - \frac{\sqrt{3}}{2} x′y′ + \frac{1}{4} (y′)^2\right) + \sqrt{3}\left(\frac{\sqrt{3}}{4} (x′)^2 + \frac{1}{2} x′y′ - \frac{\sqrt{3}}{4} (y′)^2\right) + 2\left(\frac{1}{4} (x′)^2 + \frac{\sqrt{3}}{2} x′y′ + \frac{3}{4} (y′)^2\right) - 6 = 0$$

$$\frac{9}{4} (x′)^2 - \frac{3\sqrt{3}}{2} x′y′ + \frac{3}{4} (y′)^2 + \frac{3}{4} (x′)^2 + \frac{\sqrt{3}}{2} x′y′ - \frac{3}{4} (y′)^2 + \frac{1}{2} (x′)^2 + \sqrt{3} x′y′ + \frac{3}{2} (y′)^2 - 6 = 0$$

$$\frac{7}{2} (x′)^2 + \frac{3}{2} (y′)^2 = 6$$

Divide by 6 to write to make the equation equal 1.

$$\frac{(x′)^2}{^{12}/_7} + \frac{(y′)^2}{4} = 1$$

Thus it is a vertical ellipse with a = 2 and b = \(\frac{2\sqrt{21}}{7}\) and center (0, 0).

Draw the rotated axis, then move a = 2 along the rotated y-axis and b = \(\frac{2\sqrt{21}}{7}\) along the rotated x-axis.

Connect the points with a nice ellipse.

Graph a Rotated Conic

Classify \(3x^2 - 2\sqrt{3} xy + y^2 - 16x - 16\sqrt{3} y = 0\) and then sketch the graph.

Solution

Compare \(3x^2 - 2\sqrt{3} xy + y^2 - 16x - 16\sqrt{3} y = 0\) to Ax² + Bxy + Cy² + Dx + Ey + F = 0 and find that A = 3, B = \(-2\sqrt{3}\), and C = 1.

Classify the conic by using the discriminant.

B² − 4AC

$$\left(-2\sqrt{3}\right)^2 – 4(3)(1) = 0$$

Since B² − 4AC = 0, it is a parabola.

Write it in standard form by first finding the angle of rotation.

$$\cot ⁡2θ = \frac{A - C}{B} = \frac{3 - 1}{-2\sqrt{3}} = -\frac{1}{\sqrt{3}}$$

$$2θ = \frac{2π}{3}$$

$$θ = \frac{π}{3}$$

Find the substitutions for x and y.

$$\begin{alignat}{2}x &= x′ \cos \frac{π}{3} &&- y′ \sin \frac{π}{3}\\ y &= x′ \sin \frac{π}{3} &&+ y′ \cos \frac{π}{3}\end{alignat}$$

$$\begin{alignat}{2}x &= \frac{1}{2} x′ &&- \frac{\sqrt{3}}{2} y′\\ y &= \frac{\sqrt{3}}{2} x′ &&+ \frac{1}{2} y′\end{alignat}$$

Substitute these into the original equation and simplify.

$$3\left(\frac{1}{2} x′ - \frac{\sqrt{3}}{2} y′\right)^2 - 2\sqrt{3}\left(\frac{1}{2} x′ - \frac{\sqrt{3}}{2} y′\right)\left(\frac{\sqrt{3}}{2} x′ + \frac{1}{2} y′\right) + \left(\frac{\sqrt{3}}{2} x′ + \frac{1}{2} y′\right)^2 - 16\left(\frac{1}{2} x′ - \frac{\sqrt{3}}{2} y′\right) - 16\sqrt{3} \left(\frac{\sqrt{3}}{2} x′ + \frac{1}{2} y′\right) = 0$$

$$3\left(\frac{1}{4} (x′)^2 - \frac{\sqrt{3}}{2} x′y′ + \frac{3}{4} (y′)^2\right) - 2\sqrt{3} \left(\frac{\sqrt{3}}{4} (x′)^2 - \frac{1}{2} x′y′ - \frac{\sqrt{3}}{4} (y′)^2\right) + \left(\frac{3}{4} (x′)^2 + \frac{\sqrt{3}}{2} x′y′ + \frac{1}{4} (y′)^2\right) - 16\left(\frac{1}{2} x′ - \frac{\sqrt{3}}{2} y′\right) - 16\sqrt{3} \left(\frac{\sqrt{3}}{2} x′ + \frac{1}{2} y′\right) = 0$$

$$\frac{3}{4} (x′)^2 - \frac{3\sqrt{3}}{2} x′y′ + \frac{9}{4} (y′)^2 - \frac{3}{2} (x′)^2 + \sqrt{3} x′y′ + \frac{3}{2} (y′)^2 + \frac{3}{4} (x′)^2 + \frac{\sqrt{3}}{2} x′y′ + \frac{1}{4} (y′)^2 - 8 x′ + 8\sqrt{3} y′ - 24 x′ - 8\sqrt{3} y′ = 0$$

$$4 (y′)^2 - 32 x′ = 0$$

$$(y′)^2 = 8x′$$

This is a horizontal parabola. Graph it by drawing the rotated axes and plotting points.

Graph a Rotated Conic without a Special Angle of Rotation

Classify 5x² + 4xy + 2y² − 16 = 0 and then sketch the graph.

Solution

Compare 5x² + 4xy + 2y² − 16 = 0 to Ax² + Bxy + Cy² + Dx + Ey + F = 0 and find that A = 5, B = 4, and C = 2.

Classify the conic by using the discriminant.

B² − 4AC

4² – 4(5)(2) = −24

Since B² − 4AC < 0, it is a ellipse.

Write it in standard form by first finding the angle of rotation.

$$\cot ⁡2θ = \frac{A - C}{B} = \frac{5 - 2}{4} = \frac{3}{4}$$

$$2θ = 53.13°$$

$$θ = 26.57°$$

This is not on the unit circle, so it will not produce a special angle. So the substitutions for x and y are usually found by using half-angle formulas to find sin θ and cos θ.

1. Find cot 2θ.

   $$\cot ⁡2θ = \frac{3}{4}$$

2. Reciprocal to find tan 2θ.

   $$\tan ⁡2θ = \frac{4}{3}$$

3. Use 1 + tan² u = sec² u to find sec 2θ. (If tan 2θ < 0, then sec 2θ < 0.)

   $$1 + \tan^2 2θ = \sec^2 2θ$$

   $$1 + \left(\frac{4}{3}\right)^2 = \sec^2 2θ$$

   $$1 + \frac{16}{9} = \sec^2 2θ$$

   $$\frac{25}{9} = \sec^2 2θ$$

   $$\sec 2θ = \frac{5}{3}$$

4. Reciprocal to find cos 2θ.

   $$\cos 2θ = \frac{3}{5}$$

5. Use the half-angle formulas to find sin θ and cos θ.

   $$\sin θ = \sqrt{\frac{1 - \cos 2θ}{2}} \text{ and } \cos θ = \sqrt{\frac{1 + \cos 2θ}{2}}$$

   $$\sin θ = \sqrt{\frac{1 - ^3/_5}{2}} \text{ and } \cos θ = \sqrt{\frac{1 + ^3/_5}{2}}$$

   $$\sin θ = \sqrt{\frac{^2/_5}{2}} \text{ and } \cos θ = \sqrt{\frac{^8/_5}{2}}$$

   $$\sin θ = \sqrt{\frac{1}{5}} \text{ and } \cos θ = \sqrt{\frac{4}{5}}$$

   $$\sin θ = \frac{\sqrt{5}}{5} \text{ and } \cos θ = \frac{2\sqrt{5}}{5}$$

Substitute these into the original equation and simplify.

$$5\left(\frac{2\sqrt{5}}{5} x′ - \frac{\sqrt{5}}{5} y′\right)^2 + 4 \left(\frac{2\sqrt{5}}{5} x′ - \frac{\sqrt{5}}{5} y′\right)\left(\frac{\sqrt{5}}{5} x′ + \frac{2\sqrt{5}}{5} y′\right) + 2\left(\frac{\sqrt{5}}{5} x′ + \frac{2\sqrt{5}}{5} y′\right)^2 - 16 = 0$$

$$5\left(\frac{4}{5} (x′)^2 - \frac{4}{5} x′y′ + \frac{1}{5} (y′)^2\right) + 4\left(\frac{2}{5} (x′)^2 + \frac{3}{5} x′y′ - \frac{2}{5} (y′)^2\right) + 2\left(\frac{1}{5} (x′)^2 + \frac{4}{5} x′y′ + \frac{4}{5} (y′)^2\right) - 16 = 0$$

$$4(x′)^2 - 4x′y′ + (y′)^2 + \frac{8}{5} (x′)^2 + \frac{12}{5} x′y′ - \frac{8}{5} (y′)^2 + \frac{2}{5} (x′)^2 + \frac{8}{5} x′y′ + \frac{8}{5} (y′)^2 - 16 = 0$$

$$6(x′)^2 + (y′)^2 = 16$$

Divide by 16 to write to make the equation equal 1.

$$\frac{3(x′)^2}{8} + \frac{(y′)^2}{16} = 1$$

Thus, it is a vertical ellipse with a = 4 and b = \(\frac{2\sqrt{6}}{3}\) and center (0, 0).

Draw the rotated axis, then move a = 4 along the rotated y-axis and b = \(\frac{2\sqrt{6}}{3}\) along the rotated x-axis.

Connect the points with a nice ellipse.

Classify Conics

For each of the following, classify the graph, use the quadratic formula to solve for y, and use a graphing utility to graph the equation.

1. x² + 6xy − 2y² + 3x = 0
2. 2x² – 2xy + y² – 2y = 0
3. 4x² − 4xy + y² − x + y = 0

Solution

1. Classify the graph using the discriminant. B² – 4AC = 6² – 4(1)(−2) = 44 > 0, so it is a hyperbola.

   To solve for y, rearrange terms in powers of y and factor.

   x² + 6xy − 2y² + 3x = 0

   −2y² + 6xy + (x² + 3x) = 0

   Now fill in the quadratic formula: \(y = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\).

   $$y = \frac{-6x ± \sqrt{(6x)^2 - 4(-2)(x^2 + 3x)}}{2(3)}$$

   Because of the ± sign, you will have to input two equations, one with + and one with −, to make the graph. Notice that the graph is shifted and (0, 0) is not the center.

   

2. Classify the graph using the discriminant. B² – 4AC = (−2)² – 4(2)(1)= −4 < 0, so it is an ellipse.

   To solve for y, rearrange the terms in powers of y and factor.

   2x² − 2xy + y² − 2y = 0

   y² + (−2xy − 2y) + 2x² = 0

   y² + (−2x − 2)y + 2x² = 0

   Now fill in the quadratic formula: \(y = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\).

   $$y = \frac{-(-2x - 2) ± \sqrt{(-2x - 2)^2 - 4(1)(2x^2)}}{2(1)}$$

   Because of the ± sign, you will have to input two equations, one with + and one with −, to make the graph. Notice that the graph is shifted and (0, 0) is not the center.

   

3. Classify the graph using the discriminant. B² – 4AC = (−4)² – 4(4)(1) = 0, so it is a parabola.

   To solve for y, rearrange the terms in powers of y and factor.

   4x² − 4xy + y² − x + y = 0

   y² + (−4xy + y²) + (4x² − x) = 0

   y² + (−4x + 1)y + (4x² − x) = 0

   Now fill in the quadratic formula: \(y = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\).

   $$y = \frac{-(-4x + 1) ± \sqrt{(-4x + 1)^2 - 4(1)(4x^2 - x)}}{2(1)}$$

   Because of the ± sign, you will have to input two equations, one with + and one with −, to make the graph. Notice that the graph is shifted and (0, 0) is not the vertex.