Pre-Calculus Question

Question

25x² − 120xy + 144y² − 156x − 65y = 0

(b) Use a rotation of axes to eliminate the xy-term. (Write an equation in XY-coordinates. Use a rotation angle that satisfies 0 ≤ 𝜑 ≤ 𝜋/2.)

Answer 1

Ax²+Bxy+Cy²+Dx+Ey+F=0

Let x = x₂cos(φ)-y₂sin(φ) and y = x₂sin(φ)+y₂cos(φ)

To eliminate the xy term,

substitute for x and y in the original equation and set the x₂y₂ coefficient to zero and solve for φ.

φ = tan^-1(B/(A-C))/2

Answer 2

1. Identify the Rotation Angle

The general form of the quadratic equation is:

Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0Ax2+Bxy+Cy2+Dx+Ey+F=0

In this case, A=25A = 25A=25, B=−120B = -120B=−120, C=144C = 144C=144, D=−156D = -156D=−156, E=−65E = -65E=−65, and F=0F = 0F=0.

To eliminate the xyxyxy-term, use the rotation formula. The angle ϕ\phiϕ of rotation is given by:

tan⁡(2ϕ)=BA−C\tan(2\phi) = \frac{B}{A - C}tan(2ϕ)=A−CB​

Substitute A=25A = 25A=25, B=−120B = -120B=−120, and C=144C = 144C=144:

tan⁡(2ϕ)=−12025−144=−120−119=120119\tan(2\phi) = \frac{-120}{25 - 144} = \frac{-120}{-119} = \frac{120}{119}tan(2ϕ)=25−144−120​=−119−120​=119120​

Now, find ϕ\phiϕ by solving:

2ϕ=arctan⁡(120119)2\phi = \arctan\left(\frac{120}{119}\right)2ϕ=arctan(119120​) ϕ=12arctan⁡(120119)\phi = \frac{1}{2} \arctan\left(\frac{120}{119}\right)ϕ=21​arctan(119120​)

2. Calculate the Rotation Angle

Using a calculator or software to find ϕ\phiϕ:

ϕ=12arctan⁡(120119)≈0.785 radians (or 45∘)\phi = \frac{1}{2} \arctan\left(\frac{120}{119}\right) \approx 0.785 \text{ radians} \text{ (or } 45^\circ \text{)}ϕ=21​arctan(119120​)≈0.785 radians (or 45∘)

3. Rotation Transformation

The rotation of axes formulas are:

X=xcos⁡ϕ−ysin⁡ϕY=xsin⁡ϕ+ycos⁡ϕ\begin{aligned} X &= x \cos \phi - y \sin \phi \\ Y &= x \sin \phi + y \cos \phi \end{aligned}XY​=xcosϕ−ysinϕ=xsinϕ+ycosϕ​

For ϕ=45∘\phi = 45^\circϕ=45∘ (or π4\frac{\pi}{4}4π​):

cos⁡(45∘)=12sin⁡(45∘)=12\begin{aligned} \cos(45^\circ) &= \frac{1}{\sqrt{2}} \\ \sin(45^\circ) &= \frac{1}{\sqrt{2}} \end{aligned}cos(45∘)sin(45∘)​=2​1​=2​1​​

Substitute into the rotation equations:

X=12(x−y)Y=12(x+y)\begin{aligned} X &= \frac{1}{\sqrt{2}} (x - y) \\ Y &= \frac{1}{\sqrt{2}} (x + y) \end{aligned}XY​=2​1​(x−y)=2​1​(x+y)​

4. Substitute into the Original Equation

Express xxx and yyy in terms of XXX and YYY:

x=12(X+Y)y=12(X−Y)\begin{aligned} x &= \frac{1}{\sqrt{2}} (X + Y) \\ y &= \frac{1}{\sqrt{2}} (X - Y) \end{aligned}xy​=2​1​(X+Y)=2​1​(X−Y)​