Answer :
To eliminate the [tex]\( xy \)[/tex]-term in the given equation [tex]\( xy + 6 = 0 \)[/tex], we need to rotate the coordinate system. Here are the detailed steps to accomplish this:
### Step 1: Identify the Rotation Angle
The rotation angle [tex]\(\theta\)[/tex] is determined using the formula:
[tex]\[ \theta = \frac{1}{2} \arctan\left(\frac{2B}{A - C}\right) \][/tex]
In the given equation [tex]\( xy + 6 = 0 \)[/tex], we identify the coefficients as follows:
- [tex]\( A = 0 \)[/tex]
- [tex]\( B = \frac{1}{2} ( coefficient \text{ of } xy \text{ term}) = 1 \times \frac{1}{2} = 0.5 \)[/tex]
- [tex]\( C = 0 \)[/tex]
So, the angle [tex]\(\theta\)[/tex] becomes:
[tex]\[ \theta = \frac{1}{2} \arctan\left(\frac{2 \cdot 1}{0 - 0}\right) = \frac{1}{2} \arctan(\infty) \][/tex]
This simplifies to:
[tex]\[ \theta = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \][/tex]
Thus, [tex]\(\theta = 45^\circ \)[/tex] or [tex]\(\theta = \frac{\pi}{4} \)[/tex].
### Step 2: Rotate the Coordinate System
The new coordinates [tex]\((u, v)\)[/tex] are related to the old coordinates [tex]\((x, y)\)[/tex] by a rotation of [tex]\( 45^\circ \)[/tex]:
[tex]\[ \begin{cases} u = x \cos(\theta) + y \sin(\theta) = \frac{1}{\sqrt{2}}(x + y) \\ v = -x \sin(\theta) + y \cos(\theta) = \frac{1}{\sqrt{2}}(-x + y) \end{cases} \][/tex]
Substitute [tex]\( \cos(\theta) = \sin(\theta) = \frac{1}{\sqrt{2}} \)[/tex]:
[tex]\[ \begin{cases} u = \frac{1}{\sqrt{2}} (x + y) \\ v = \frac{1}{\sqrt{2}} (-x + y) \end{cases} \][/tex]
### Step 3: Substitute [tex]\(u\)[/tex] and [tex]\(v\)[/tex] into the Original Equation
Substitute the expressions of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] into [tex]\(xy + 6 = 0\)[/tex]:
[tex]\[ \begin{cases} x = \frac{u - v}{\sqrt{2}} \\ y = \frac{u + v}{\sqrt{2}} \end{cases} \][/tex]
Substitute these into the given equation:
[tex]\[ \left( \frac{u - v}{\sqrt{2}} \right) \left( \frac{u + v}{\sqrt{2}} \right) + 6 = 0 \][/tex]
[tex]\[ \frac{1}{2} (u^2 - v^2) + 6 = 0 \][/tex]
Multiply both sides by 2 to simplify:
[tex]\[ u^2 - v^2 = -12 \][/tex]
This is the equation of a hyperbola in the rotated coordinate system.
### Step 4: Sketch the Graph
The given equation in the new coordinate system is:
[tex]\[ u^2 - v^2 = -12 \][/tex]
To sketch the graph, reverse the transformations:
- [tex]\( u = \frac{x + y}{\sqrt{2}} \)[/tex]
- [tex]\( v = \frac{-x + y}{\sqrt{2}} \)[/tex]
Graph the hyperbola in the rotated system first, then transform back to the original coordinate system:
[tex]\[ xy + 6 = 0 \rightarrow xy = -6 \][/tex]
The graph of [tex]\( xy = -6 \)[/tex] is a rectangular hyperbola with asymptotes along the lines [tex]\(x = 0\)[/tex] and [tex]\(y = 0\)[/tex].
### Final Answer
The rotated equation in standard form is:
[tex]\[ u^2 - v^2 = -12 \][/tex]
The graph is a hyperbola with asymptotes aligned with [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex], opening along the coordinate axes.
### Step 1: Identify the Rotation Angle
The rotation angle [tex]\(\theta\)[/tex] is determined using the formula:
[tex]\[ \theta = \frac{1}{2} \arctan\left(\frac{2B}{A - C}\right) \][/tex]
In the given equation [tex]\( xy + 6 = 0 \)[/tex], we identify the coefficients as follows:
- [tex]\( A = 0 \)[/tex]
- [tex]\( B = \frac{1}{2} ( coefficient \text{ of } xy \text{ term}) = 1 \times \frac{1}{2} = 0.5 \)[/tex]
- [tex]\( C = 0 \)[/tex]
So, the angle [tex]\(\theta\)[/tex] becomes:
[tex]\[ \theta = \frac{1}{2} \arctan\left(\frac{2 \cdot 1}{0 - 0}\right) = \frac{1}{2} \arctan(\infty) \][/tex]
This simplifies to:
[tex]\[ \theta = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \][/tex]
Thus, [tex]\(\theta = 45^\circ \)[/tex] or [tex]\(\theta = \frac{\pi}{4} \)[/tex].
### Step 2: Rotate the Coordinate System
The new coordinates [tex]\((u, v)\)[/tex] are related to the old coordinates [tex]\((x, y)\)[/tex] by a rotation of [tex]\( 45^\circ \)[/tex]:
[tex]\[ \begin{cases} u = x \cos(\theta) + y \sin(\theta) = \frac{1}{\sqrt{2}}(x + y) \\ v = -x \sin(\theta) + y \cos(\theta) = \frac{1}{\sqrt{2}}(-x + y) \end{cases} \][/tex]
Substitute [tex]\( \cos(\theta) = \sin(\theta) = \frac{1}{\sqrt{2}} \)[/tex]:
[tex]\[ \begin{cases} u = \frac{1}{\sqrt{2}} (x + y) \\ v = \frac{1}{\sqrt{2}} (-x + y) \end{cases} \][/tex]
### Step 3: Substitute [tex]\(u\)[/tex] and [tex]\(v\)[/tex] into the Original Equation
Substitute the expressions of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] into [tex]\(xy + 6 = 0\)[/tex]:
[tex]\[ \begin{cases} x = \frac{u - v}{\sqrt{2}} \\ y = \frac{u + v}{\sqrt{2}} \end{cases} \][/tex]
Substitute these into the given equation:
[tex]\[ \left( \frac{u - v}{\sqrt{2}} \right) \left( \frac{u + v}{\sqrt{2}} \right) + 6 = 0 \][/tex]
[tex]\[ \frac{1}{2} (u^2 - v^2) + 6 = 0 \][/tex]
Multiply both sides by 2 to simplify:
[tex]\[ u^2 - v^2 = -12 \][/tex]
This is the equation of a hyperbola in the rotated coordinate system.
### Step 4: Sketch the Graph
The given equation in the new coordinate system is:
[tex]\[ u^2 - v^2 = -12 \][/tex]
To sketch the graph, reverse the transformations:
- [tex]\( u = \frac{x + y}{\sqrt{2}} \)[/tex]
- [tex]\( v = \frac{-x + y}{\sqrt{2}} \)[/tex]
Graph the hyperbola in the rotated system first, then transform back to the original coordinate system:
[tex]\[ xy + 6 = 0 \rightarrow xy = -6 \][/tex]
The graph of [tex]\( xy = -6 \)[/tex] is a rectangular hyperbola with asymptotes along the lines [tex]\(x = 0\)[/tex] and [tex]\(y = 0\)[/tex].
### Final Answer
The rotated equation in standard form is:
[tex]\[ u^2 - v^2 = -12 \][/tex]
The graph is a hyperbola with asymptotes aligned with [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex], opening along the coordinate axes.
