10.4 Rotation of axes  (Page 4/8)

First, we find $\cot \left(2\theta \right).$

Because $\cot \left(2\theta \right)=\frac{5}{12},$ we can draw a reference triangle as in [link] .

Thus, the hypotenuse is

Next, we find $\sin \theta$ and $\cos \theta .$ We will use half-angle identities.

Now we find $x$ and $y\text{.\hspace{0.17em}}$

and

Now we substitute $x=\frac{3x^{\prime }-2y^{\prime }}{\sqrt{13}}$ and $y=\frac{2x^{\prime }+3y^{\prime }}{\sqrt{13}}$ into $x^2+12xy-4y^2=30.$

[link] shows the graph of the hyperbola $\frac{{x^{\prime }}^2}{6}-\frac{4{y^{\prime }}^2}{15}=1.$

1. Let’s begin by determining $A,B,$ and $C.$
   $$\underset{A}{\underbrace{5}}{x}^2+\underset{B}{\underbrace{2\sqrt{3}}}xy+\underset{C}{\underbrace{2}}{y}^2-5=0$$

   Now, we find the discriminant.

   $$\begin{array}{l}{B}^2-4AC={\left(2\sqrt{3}\right)}^2-4(5)(2)\hfill \\ =4(3)-40\hfill \\ =12-40\hfill \\ =-28<0\hfill \end{array}$$

   Therefore, $5x^2+2\sqrt{3}xy+2y^2-5=0$ represents an ellipse.

2. Again, let’s begin by determining $A,B,$ and $C.$
   $$\underset{A}{\underbrace{5}}{x}^2+\underset{B}{\underbrace{2\sqrt{3}}}xy+\underset{C}{\underbrace{12}}{y}^2-5=0$$

   Now, we find the discriminant.

   $$\begin{array}{l}{B}^2-4AC={\left(2\sqrt{3}\right)}^2-4(5)(12)\hfill \\ =4(3)-240\hfill \\ =12-240\hfill \\ =-228<0\hfill \end{array}$$

   Therefore, $5x^2+2\sqrt{3}xy+12y^2-5=0$ represents an ellipse.