Math / Geometry

QuestionRotate the axes to eliminate the $x y$ -term in the equation. Then write the equation in standard form. (Use $x p$ for $x^{\prime}$ and $y p$ for $y^{\prime}$ in your answer. Rotate the coordinate axes through an angle $\theta$ with $0 \leq \theta \leq \pi / 2$ .) $2 x^{2}+x y+2 y^{2}-8=0$

Studdy Solution

STEP 1

What is this asking? We need to rotate the coordinate axes to get rid of the pesky $xy$ term in the given equation and then rewrite the equation in a nice, standard form. Watch out! Remember, rotating the axes changes how we describe the points, not the actual shape of the curve!
Also, don't forget the restriction on $\theta$ !

STEP 2

1. Find the rotation angle.
2. Substitute and simplify.
3. Write in standard form.

STEP 3

Alright, let's find that perfect rotation angle, $\theta$ !
The formula for the angle of rotation to eliminate the $xy$ term is given by $\cot(2\theta) = \frac{A - C}{B}$ , where $A$ , $B$ , and $C$ are the coefficients of $x^2$ , $xy$ , and $y^2$ respectively.

STEP 4

In our equation, $2x^2 + xy + 2y^2 - 8 = 0$ , we have $A = 2$ , $B = 1$ , and $C = 2$ .
So, $\cot(2\theta) = \frac{2 - 2}{1} = \frac{0}{1} = 0$ .

STEP 5

Now, we need to find a $\theta$ between $0$ and $\pi/2$ such that $\cot(2\theta) = 0$ .
This happens when $2\theta = \frac{\pi}{2}$ , which means $\theta = \frac{\pi}{4}$ .
Awesome! We've got our rotation angle: $\theta = \frac{\pi}{4}$ .

STEP 6

Now for the fun part!
We'll use the rotation formulas: $x = xp \cos\theta - yp \sin\theta$ $y = xp \sin\theta + yp \cos\theta$ With $\theta = \frac{\pi}{4}$ , we have $\cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ .
Substituting these values, we get: $x = \frac{\sqrt{2}}{2}(xp - yp)$ $y = \frac{\sqrt{2}}{2}(xp + yp)$

STEP 7

Let's plug these expressions for $x$ and $y$ into our original equation $2x^2 + xy + 2y^2 - 8 = 0$ : $2\left(\frac{\sqrt{2}}{2}(xp - yp)\right)^2 + \left(\frac{\sqrt{2}}{2}(xp - yp)\right)\left(\frac{\sqrt{2}}{2}(xp + yp)\right) + 2\left(\frac{\sqrt{2}}{2}(xp + yp)\right)^2 - 8 = 0$

STEP 8

Time to simplify!
After expanding and combining like terms, we get: $(xp)^2 - 2(xp)(yp) + (yp)^2 + \frac{1}{2}((xp)^2 - (yp)^2) + (xp)^2 + 2(xp)(yp) + (yp)^2 - 8 = 0$ $\frac{5}{2}(xp)^2 + \frac{3}{2}(yp)^2 - 8 = 0$

STEP 9

Almost there!
To get the standard form, we want the equation to equal 1.
So, let's move the 8 to the other side and divide everything by 8: $\frac{5}{16}(xp)^2 + \frac{3}{16}(yp)^2 = 1$

STEP 10

Our equation in standard form is $\frac{5}{16}(xp)^2 + \frac{3}{16}(yp)^2 = 1$ .

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QuestionRotate the axes to eliminate the $x y$ -term in the equation. Then write the equation in standard form. (Use $x p$ for $x^{\prime}$ and $y p$ for $y^{\prime}$ in your answer. Rotate the coordinate axes through an angle $\theta$ with $0 \leq \theta \leq \pi / 2$ .) $2 x^{2}+x y+2 y^{2}-8=0$

Studdy Solution

STEP 1

STEP 2

1. Find the rotation angle.
2. Substitute and simplify.
3. Write in standard form.

STEP 3

Alright, let's find that perfect rotation angle, $\theta$ !
The formula for the angle of rotation to eliminate the $xy$ term is given by $\cot(2\theta) = \frac{A - C}{B}$ , where $A$ , $B$ , and $C$ are the coefficients of $x^2$ , $xy$ , and $y^2$ respectively.

STEP 4

In our equation, $2x^2 + xy + 2y^2 - 8 = 0$ , we have $A = 2$ , $B = 1$ , and $C = 2$ .
So, $\cot(2\theta) = \frac{2 - 2}{1} = \frac{0}{1} = 0$ .

STEP 5

Now, we need to find a $\theta$ between $0$ and $\pi/2$ such that $\cot(2\theta) = 0$ .
This happens when $2\theta = \frac{\pi}{2}$ , which means $\theta = \frac{\pi}{4}$ .
Awesome! We've got our rotation angle: $\theta = \frac{\pi}{4}$ .

STEP 6

STEP 7

Let's plug these expressions for $x$ and $y$ into our original equation $2x^2 + xy + 2y^2 - 8 = 0$ : $2\left(\frac{\sqrt{2}}{2}(xp - yp)\right)^2 + \left(\frac{\sqrt{2}}{2}(xp - yp)\right)\left(\frac{\sqrt{2}}{2}(xp + yp)\right) + 2\left(\frac{\sqrt{2}}{2}(xp + yp)\right)^2 - 8 = 0$

STEP 8

Time to simplify!
After expanding and combining like terms, we get: $(xp)^2 - 2(xp)(yp) + (yp)^2 + \frac{1}{2}((xp)^2 - (yp)^2) + (xp)^2 + 2(xp)(yp) + (yp)^2 - 8 = 0$ $\frac{5}{2}(xp)^2 + \frac{3}{2}(yp)^2 - 8 = 0$

STEP 9

Almost there!
To get the standard form, we want the equation to equal 1.
So, let's move the 8 to the other side and divide everything by 8: $\frac{5}{16}(xp)^2 + \frac{3}{16}(yp)^2 = 1$

STEP 10

Our equation in standard form is $\frac{5}{16}(xp)^2 + \frac{3}{16}(yp)^2 = 1$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the rotation angle.2. Substitute and simplify.3. Write in standard form.

STEP 3

STEP 4

In our equation, 2x2+xy+2y2−8=02x^2 + xy + 2y^2 - 8 = 02x2+xy+2y2−8=0, we have A=2A = 2A=2, B=1B = 1B=1, and C=2C = 2C=2.So, cot⁡(2θ)=2−21=01=0\cot(2\theta) = \frac{2 - 2}{1} = \frac{0}{1} = 0cot(2θ)=12−2​=10​=0.

STEP 5

STEP 6

STEP 7

STEP 8

STEP 9

Almost there!To get the standard form, we want the equation to equal 1.So, let's move the 8 to the other side and divide everything by 8: 516(xp)2+316(yp)2=1\frac{5}{16}(xp)^2 + \frac{3}{16}(yp)^2 = 1165​(xp)2+163​(yp)2=1

STEP 10

Our equation in standard form is 516(xp)2+316(yp)2=1\frac{5}{16}(xp)^2 + \frac{3}{16}(yp)^2 = 1165​(xp)2+163​(yp)2=1.

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1. Find the rotation angle.
2. Substitute and simplify.
3. Write in standard form.

In our equation, $2x^2 + xy + 2y^2 - 8 = 0$ , we have $A = 2$ , $B = 1$ , and $C = 2$ .
So, $\cot(2\theta) = \frac{2 - 2}{1} = \frac{0}{1} = 0$ .

Almost there!
To get the standard form, we want the equation to equal 1.
So, let's move the 8 to the other side and divide everything by 8: $\frac{5}{16}(xp)^2 + \frac{3}{16}(yp)^2 = 1$

Our equation in standard form is $\frac{5}{16}(xp)^2 + \frac{3}{16}(yp)^2 = 1$ .