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To solve the problem of rotating the axes to eliminate the [tex]\(xy\)[/tex]-term in the equation [tex]\(xy + 4 = 0\)[/tex] and then writing the equation in standard form, we can follow a series of steps:
### Step 1: Identifying the Coefficients
First, we identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(f\)[/tex] from the given equation:
[tex]\[ xy + 4 = 0 \][/tex]
This corresponds to:
[tex]\[ 0x^2 + xy + 0y^2 + 4 = 0 \][/tex]
From this, we see that:
[tex]\[ a = 0, \quad b = 1, \quad c = 0, \quad f = 4 \][/tex]
### Step 2: Finding the Angle of Rotation
To eliminate the [tex]\(xy\)[/tex]-term, we must find the angle [tex]\(\theta\)[/tex] that satisfies the following condition:
[tex]\[ \tan(2\theta) = \frac{b}{a - c} \][/tex]
Substituting our values:
[tex]\[ \tan(2\theta) = \frac{1}{0 - 0} = \frac{1}{0} \quad \text{(since } 0 - 0 = 0 \text{ and the division by zero leads to an infinite value)} \][/tex]
For [tex]\(\tan(2\theta)\)[/tex] to be infinite, [tex]\(2\theta = \frac{\pi}{2}\)[/tex]. Therefore:
[tex]\[ \theta = \frac{\pi}{4} \][/tex]
### Step 3: Calculating the New Coefficients in the Rotated Coordinate System
With [tex]\(\theta = \frac{\pi}{4}\)[/tex], we can find the new coefficients [tex]\(a'\)[/tex], [tex]\(b'\)[/tex], and [tex]\(c'\)[/tex]:
[tex]\[ a' = \frac{a + c}{2} + \frac{a - c}{2} \cos(2\theta) + \frac{b}{2} \sin(2\theta) \][/tex]
[tex]\[ c' = \frac{a + c}{2} - \frac{a - c}{2} \cos(2\theta) - \frac{b}{2} \sin(2\theta) \][/tex]
[tex]\[ b' = 0 \quad (\text{since the purpose of rotation is to eliminate the } xy \text{-term}) \][/tex]
Let's calculate each of these new coefficients:
[tex]\[ a' = \frac{0+0}{2} + \frac{0-0}{2} \cos(\frac{\pi}{2}) + \frac{1}{2} \sin(\frac{\pi}{2}) \][/tex]
Since [tex]\(\cos(\frac{\pi}{2}) = 0\)[/tex] and [tex]\(\sin(\frac{\pi}{2}) = 1\)[/tex]:
[tex]\[ a' = 0 + 0 + \frac{1}{2} \times 1 = \frac{1}{2} = 0.5 \][/tex]
[tex]\[ c' = \frac{0+0}{2} - \frac{0-0}{2} \cos(\frac{\pi}{2}) - \frac{1}{2} \sin(\frac{\pi}{2}) \][/tex]
[tex]\[ c' = 0 - 0 - \frac{1}{2} \times 1 = -\frac{1}{2} = -0.5 \][/tex]
Finally, the [tex]\(f\)[/tex] term remains unchanged:
[tex]\[ f' = 4 \][/tex]
### Step 4: Writing the Equation in Standard Form
Therefore, in the rotated coordinate system, the equation becomes:
[tex]\[ 0.5(x')^2 - 0.5(y')^2 + 4 = 0 \][/tex]
Hence, the transformed equation without the [tex]\(xy\)[/tex]-term is:
[tex]\[ 0.5(xp)^2 - 0.5(yp)^2 + 4 = 0 \][/tex]
This is the standard form of the equation after rotating through an angle of [tex]\(\theta = \frac{\pi}{4}\)[/tex].
### Step 1: Identifying the Coefficients
First, we identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(f\)[/tex] from the given equation:
[tex]\[ xy + 4 = 0 \][/tex]
This corresponds to:
[tex]\[ 0x^2 + xy + 0y^2 + 4 = 0 \][/tex]
From this, we see that:
[tex]\[ a = 0, \quad b = 1, \quad c = 0, \quad f = 4 \][/tex]
### Step 2: Finding the Angle of Rotation
To eliminate the [tex]\(xy\)[/tex]-term, we must find the angle [tex]\(\theta\)[/tex] that satisfies the following condition:
[tex]\[ \tan(2\theta) = \frac{b}{a - c} \][/tex]
Substituting our values:
[tex]\[ \tan(2\theta) = \frac{1}{0 - 0} = \frac{1}{0} \quad \text{(since } 0 - 0 = 0 \text{ and the division by zero leads to an infinite value)} \][/tex]
For [tex]\(\tan(2\theta)\)[/tex] to be infinite, [tex]\(2\theta = \frac{\pi}{2}\)[/tex]. Therefore:
[tex]\[ \theta = \frac{\pi}{4} \][/tex]
### Step 3: Calculating the New Coefficients in the Rotated Coordinate System
With [tex]\(\theta = \frac{\pi}{4}\)[/tex], we can find the new coefficients [tex]\(a'\)[/tex], [tex]\(b'\)[/tex], and [tex]\(c'\)[/tex]:
[tex]\[ a' = \frac{a + c}{2} + \frac{a - c}{2} \cos(2\theta) + \frac{b}{2} \sin(2\theta) \][/tex]
[tex]\[ c' = \frac{a + c}{2} - \frac{a - c}{2} \cos(2\theta) - \frac{b}{2} \sin(2\theta) \][/tex]
[tex]\[ b' = 0 \quad (\text{since the purpose of rotation is to eliminate the } xy \text{-term}) \][/tex]
Let's calculate each of these new coefficients:
[tex]\[ a' = \frac{0+0}{2} + \frac{0-0}{2} \cos(\frac{\pi}{2}) + \frac{1}{2} \sin(\frac{\pi}{2}) \][/tex]
Since [tex]\(\cos(\frac{\pi}{2}) = 0\)[/tex] and [tex]\(\sin(\frac{\pi}{2}) = 1\)[/tex]:
[tex]\[ a' = 0 + 0 + \frac{1}{2} \times 1 = \frac{1}{2} = 0.5 \][/tex]
[tex]\[ c' = \frac{0+0}{2} - \frac{0-0}{2} \cos(\frac{\pi}{2}) - \frac{1}{2} \sin(\frac{\pi}{2}) \][/tex]
[tex]\[ c' = 0 - 0 - \frac{1}{2} \times 1 = -\frac{1}{2} = -0.5 \][/tex]
Finally, the [tex]\(f\)[/tex] term remains unchanged:
[tex]\[ f' = 4 \][/tex]
### Step 4: Writing the Equation in Standard Form
Therefore, in the rotated coordinate system, the equation becomes:
[tex]\[ 0.5(x')^2 - 0.5(y')^2 + 4 = 0 \][/tex]
Hence, the transformed equation without the [tex]\(xy\)[/tex]-term is:
[tex]\[ 0.5(xp)^2 - 0.5(yp)^2 + 4 = 0 \][/tex]
This is the standard form of the equation after rotating through an angle of [tex]\(\theta = \frac{\pi}{4}\)[/tex].
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