Solved on Nov 18, 2023

Rewrite rotation formulas to eliminate $x'y'$ -term in equation $x^2 + 16xy + y^2 - 3 = 0$ .

STEP 1

Assumptions1. The given equation is $x^{}+16 x y+y^{}-3=0$ . We want to eliminate the $x'y'$ term in the rotated system.

STEP 2

In a rotated coordinate system, the coordinates $(x, y)$ can be transformed to $(x', y')$ using the rotation formulas $x = x' \cos(\theta) - y' \sin(\theta)$ $y = x' \sin(\theta) + y' \cos(\theta)$ where $\theta$ is the angle of rotation.

STEP 3

Substitute the rotation formulas into the given equation $x^{2}+16 x y+y^{2}-3=0$ to get $(x' \cos(\theta) - y' \sin(\theta))^2 +16(x' \cos(\theta) - y' \sin(\theta))(x' \sin(\theta) + y' \cos(\theta)) + (x' \sin(\theta) + y' \cos(\theta))^2 -3 =0$

STEP 4

We want to eliminate the $x'y'$ term in the rotated system. This will happen when the coefficient of $x'y'$ in the expanded equation is zero.The coefficient of $x'y'$ is obtained by multiplying the coefficients of $x' y'$ in the terms $x'^2$ , $x'y'$ , and $y'^2$ .
In the expanded equation, the coefficient of $x'y'$ is $16\cos(\theta)\sin(\theta) -2\sin(\theta)\cos(\theta)$ .

STEP 5

Set the coefficient of $x'y'$ to zero and solve for $\theta$ $16\cos(\theta)\sin(\theta) -2\sin(\theta)\cos(\theta) =0$

STEP 6

Factor out $\sin(\theta)\cos(\theta)$ to simplify the equation $\sin(\theta)\cos(\theta)(16 -2) =0$

STEP 7

olve for $\theta$ $\sin(\theta)\cos(\theta) =0$ This equation is satisfied when $\theta =0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ .

STEP 8

However, a rotation by $0$ or $\pi$ radians would not change the system, and a rotation by $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ radians would merely swap the $x$ and $y$ axes. Therefore, none of these solutions eliminate the $x'y'$ term.

STEP 9

The problem lies in the fact that we have a term with $x'y'$ in the equation, which cannot be eliminated by a simple rotation. This is because the given equation represents a hyperbola, which always has an $x'y'$ term in its general form.

STEP 10

To eliminate the $x'y'$ term, we need to rotate the axes by an angle $\theta$ such that the axes align with the symmetry axes of the hyperbola. This angle $\theta$ can be found by calculating the angle of the lines of symmetry of the hyperbola.

STEP 11

The angle $\theta$ is given by\theta = \frac{}{} \arctan\left(\frac{ab}{a^ - b^}\right)where $a$ and $b$ are the coefficients of x^ and y^ respectively.

STEP 12

Substitute $a =$ and $b =$ into the formula to calculate $\theta$ $\theta = \frac{}{2} \arctan\left(\frac{2()()}{^2 -^2}\right)$

STEP 13

implify the equation to find $\theta$ $\theta = \frac{}{2} \arctan(\infty)$

STEP 14

The $\arctan(\infty)$ is $\frac{\pi}{2}$ , so $\theta = \frac{}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$

STEP 15

Therefore, a rotation of $\frac{\pi}{4}$ radians (or45 degrees) will align the axes with the symmetry axes of the hyperbola and eliminate the $x'y'$ term.
The rotation formulas that will transform the given equation into a form without an $x'y'$ term are $x' = x \cos\left(\frac{\pi}{4}\right) - y \sin\left(\frac{\pi}{4}\right)$ $y' = x \sin\left(\frac{\pi}{4}\right) + y \cos\left(\frac{\pi}{4}\right)$

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Rewrite rotation formulas to eliminate $x'y'$ -term in equation $x^2 + 16xy + y^2 - 3 = 0$ .

STEP 1

Assumptions1. The given equation is $x^{}+16 x y+y^{}-3=0$ . We want to eliminate the $x'y'$ term in the rotated system.

STEP 2

In a rotated coordinate system, the coordinates $(x, y)$ can be transformed to $(x', y')$ using the rotation formulas $x = x' \cos(\theta) - y' \sin(\theta)$ $y = x' \sin(\theta) + y' \cos(\theta)$ where $\theta$ is the angle of rotation.

STEP 3

Substitute the rotation formulas into the given equation $x^{2}+16 x y+y^{2}-3=0$ to get $(x' \cos(\theta) - y' \sin(\theta))^2 +16(x' \cos(\theta) - y' \sin(\theta))(x' \sin(\theta) + y' \cos(\theta)) + (x' \sin(\theta) + y' \cos(\theta))^2 -3 =0$

STEP 4

STEP 5

Set the coefficient of $x'y'$ to zero and solve for $\theta$ $16\cos(\theta)\sin(\theta) -2\sin(\theta)\cos(\theta) =0$

STEP 6

Factor out $\sin(\theta)\cos(\theta)$ to simplify the equation $\sin(\theta)\cos(\theta)(16 -2) =0$

STEP 7

olve for $\theta$ $\sin(\theta)\cos(\theta) =0$ This equation is satisfied when $\theta =0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ .

STEP 8

However, a rotation by $0$ or $\pi$ radians would not change the system, and a rotation by $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ radians would merely swap the $x$ and $y$ axes. Therefore, none of these solutions eliminate the $x'y'$ term.

STEP 9

The problem lies in the fact that we have a term with $x'y'$ in the equation, which cannot be eliminated by a simple rotation. This is because the given equation represents a hyperbola, which always has an $x'y'$ term in its general form.

STEP 10

To eliminate the $x'y'$ term, we need to rotate the axes by an angle $\theta$ such that the axes align with the symmetry axes of the hyperbola. This angle $\theta$ can be found by calculating the angle of the lines of symmetry of the hyperbola.

STEP 11

The angle $\theta$ is given by\theta = \frac{}{} \arctan\left(\frac{ab}{a^ - b^}\right)where $a$ and $b$ are the coefficients of x^ and y^ respectively.

STEP 12

Substitute $a =$ and $b =$ into the formula to calculate $\theta$ $\theta = \frac{}{2} \arctan\left(\frac{2()()}{^2 -^2}\right)$

STEP 13

implify the equation to find $\theta$ $\theta = \frac{}{2} \arctan(\infty)$

STEP 14

The $\arctan(\infty)$ is $\frac{\pi}{2}$ , so $\theta = \frac{}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$

STEP 15

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Rewrite rotation formulas to eliminate x′y′x'y'x′y′-term in equation x2+16xy+y2−3=0x^2 + 16xy + y^2 - 3 = 0x2+16xy+y2−3=0.

STEP 1

Assumptions1. The given equation is x+16xy+y−3=0x^{}+16 x y+y^{}-3=0x+16xy+y−3=0 . We want to eliminate the x′y′x'y'x′y′ term in the rotated system.

STEP 2

STEP 3

STEP 4

STEP 5

Set the coefficient of x′y′x'y'x′y′ to zero and solve for θ\thetaθ16cos⁡(θ)sin⁡(θ)−2sin⁡(θ)cos⁡(θ)=016\cos(\theta)\sin(\theta) -2\sin(\theta)\cos(\theta) =016cos(θ)sin(θ)−2sin(θ)cos(θ)=0

STEP 6

Factor out sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta)sin(θ)cos(θ) to simplify the equationsin⁡(θ)cos⁡(θ)(16−2)=0\sin(\theta)\cos(\theta)(16 -2) =0sin(θ)cos(θ)(16−2)=0

STEP 7

olve for θ\thetaθsin⁡(θ)cos⁡(θ)=0\sin(\theta)\cos(\theta) =0sin(θ)cos(θ)=0This equation is satisfied when θ=0,π2,π,3π2\theta =0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}θ=0,2π​,π,23π​.

STEP 8

However, a rotation by 000 or π\piπ radians would not change the system, and a rotation by π2\frac{\pi}{2}2π​ or 3π2\frac{3\pi}{2}23π​ radians would merely swap the xxx and yyy axes. Therefore, none of these solutions eliminate the x′y′x'y'x′y′ term.

STEP 9

The problem lies in the fact that we have a term with x′y′x'y'x′y′ in the equation, which cannot be eliminated by a simple rotation. This is because the given equation represents a hyperbola, which always has an x′y′x'y'x′y′ term in its general form.

STEP 10

To eliminate the x′y′x'y'x′y′ term, we need to rotate the axes by an angle θ\thetaθ such that the axes align with the symmetry axes of the hyperbola. This angle θ\thetaθ can be found by calculating the angle of the lines of symmetry of the hyperbola.

STEP 11

The angle θ\thetaθ is given by\theta = \frac{}{} \arctan\left(\frac{ab}{a^ - b^}\right)where aaa and bbb are the coefficients of x^ and y^ respectively.

STEP 12

Substitute a=a =a= and b=b =b= into the formula to calculate θ\thetaθθ=2arctan⁡(2()()2−2)\theta = \frac{}{2} \arctan\left(\frac{2()()}{^2 -^2}\right)θ=2​arctan(2−22()()​)

STEP 13

implify the equation to find θ\thetaθθ=2arctan⁡(∞)\theta = \frac{}{2} \arctan(\infty)θ=2​arctan(∞)

STEP 14

The arctan⁡(∞)\arctan(\infty)arctan(∞) is π2\frac{\pi}{2}2π​, soθ=2⋅π2=π4\theta = \frac{}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}θ=2​⋅2π​=4π​

STEP 15

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Rewrite rotation formulas to eliminate $x'y'$ -term in equation $x^2 + 16xy + y^2 - 3 = 0$ .

Assumptions1. The given equation is $x^{}+16 x y+y^{}-3=0$ . We want to eliminate the $x'y'$ term in the rotated system.

Set the coefficient of $x'y'$ to zero and solve for $\theta$ $16\cos(\theta)\sin(\theta) -2\sin(\theta)\cos(\theta) =0$

Factor out $\sin(\theta)\cos(\theta)$ to simplify the equation $\sin(\theta)\cos(\theta)(16 -2) =0$

olve for $\theta$ $\sin(\theta)\cos(\theta) =0$ This equation is satisfied when $\theta =0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ .

However, a rotation by $0$ or $\pi$ radians would not change the system, and a rotation by $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ radians would merely swap the $x$ and $y$ axes. Therefore, none of these solutions eliminate the $x'y'$ term.

The problem lies in the fact that we have a term with $x'y'$ in the equation, which cannot be eliminated by a simple rotation. This is because the given equation represents a hyperbola, which always has an $x'y'$ term in its general form.

To eliminate the $x'y'$ term, we need to rotate the axes by an angle $\theta$ such that the axes align with the symmetry axes of the hyperbola. This angle $\theta$ can be found by calculating the angle of the lines of symmetry of the hyperbola.

The angle $\theta$ is given by\theta = \frac{}{} \arctan\left(\frac{ab}{a^ - b^}\right)where $a$ and $b$ are the coefficients of x^ and y^ respectively.

Substitute $a =$ and $b =$ into the formula to calculate $\theta$ $\theta = \frac{}{2} \arctan\left(\frac{2()()}{^2 -^2}\right)$

implify the equation to find $\theta$ $\theta = \frac{}{2} \arctan(\infty)$

The $\arctan(\infty)$ is $\frac{\pi}{2}$ , so $\theta = \frac{}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$