# 8.4 Rotation of axes  (Page 3/8)

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## Finding a new representation of an equation after rotating through a given angle

Find a new representation of the equation $\text{\hspace{0.17em}}2{x}^{2}-xy+2{y}^{2}-30=0\text{\hspace{0.17em}}$ after rotating through an angle of $\text{\hspace{0.17em}}\theta =45°.$

Find $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y,$ where and

Because $\text{\hspace{0.17em}}\theta =45°,$

$\begin{array}{l}\hfill \\ x={x}^{\prime }\mathrm{cos}\left(45°\right)-{y}^{\prime }\mathrm{sin}\left(45°\right)\hfill \\ x={x}^{\prime }\left(\frac{1}{\sqrt{2}}\right)-{y}^{\prime }\left(\frac{1}{\sqrt{2}}\right)\hfill \\ x=\frac{{x}^{\prime }-{y}^{\prime }}{\sqrt{2}}\hfill \end{array}$

and

$\begin{array}{l}\\ \begin{array}{l}y={x}^{\prime }\mathrm{sin}\left(45°\right)+{y}^{\prime }\mathrm{cos}\left(45°\right)\hfill \\ y={x}^{\prime }\left(\frac{1}{\sqrt{2}}\right)+{y}^{\prime }\left(\frac{1}{\sqrt{2}}\right)\hfill \\ y=\frac{{x}^{\prime }+{y}^{\prime }}{\sqrt{2}}\hfill \end{array}\end{array}$

Substitute $\text{\hspace{0.17em}}x={x}^{\prime }\mathrm{cos}\theta -{y}^{\prime }\mathrm{sin}\theta \text{\hspace{0.17em}}$ and into $\text{\hspace{0.17em}}2{x}^{2}-xy+2{y}^{2}-30=0.$

$2{\left(\frac{{x}^{\prime }-{y}^{\prime }}{\sqrt{2}}\right)}^{2}-\left(\frac{{x}^{\prime }-{y}^{\prime }}{\sqrt{2}}\right)\left(\frac{{x}^{\prime }+{y}^{\prime }}{\sqrt{2}}\right)+2{\left(\frac{{x}^{\prime }+{y}^{\prime }}{\sqrt{2}}\right)}^{2}-30=0$

Simplify.

Write the equations with $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime }\text{\hspace{0.17em}}$ in the standard form.

$\frac{{{x}^{\prime }}^{2}}{20}+\frac{{{y}^{\prime }}^{2}}{12}=1$

This equation is an ellipse. [link] shows the graph.

## Writing equations of rotated conics in standard form

Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form $\text{\hspace{0.17em}}A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0\text{\hspace{0.17em}}$ into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime }\text{\hspace{0.17em}}$ coordinate system without the $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ term, by rotating the axes by a measure of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ that satisfies

$\mathrm{cot}\left(2\theta \right)=\frac{A-C}{B}$

We have learned already that any conic may be represented by the second degree equation

$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$

where $\text{\hspace{0.17em}}A,B,$ and $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ are not all zero. However, if $\text{\hspace{0.17em}}B\ne 0,$ then we have an $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}\mathrm{cot}\left(2\theta \right)=\frac{A-C}{B}.$

• If $\text{\hspace{0.17em}}\mathrm{cot}\left(2\theta \right)>0,$ then $\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ is in the first quadrant, and $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is between $\text{\hspace{0.17em}}\left(0°,45°\right).$
• If $\text{\hspace{0.17em}}\mathrm{cot}\left(2\theta \right)<0,$ then $\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ is in the second quadrant, and $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is between $\text{\hspace{0.17em}}\left(45°,90°\right).$
• If $\text{\hspace{0.17em}}A=C,$ then $\text{\hspace{0.17em}}\theta =45°.$

Given an equation for a conic in the $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ system, rewrite the equation without the $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ term in terms of $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime },$ where the $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime }\text{\hspace{0.17em}}$ axes are rotations of the standard axes by $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ degrees.

1. Find $\text{\hspace{0.17em}}\mathrm{cot}\left(2\theta \right).$
2. Find and
3. Substitute and into and
4. Substitute the expression for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ into in the given equation, and then simplify.
5. Write the equations with $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime }\text{\hspace{0.17em}}$ in the standard form with respect to the rotated axes.

## Rewriting an equation with respect to the x′ And y′ Axes without the x′y′ Term

Rewrite the equation $\text{\hspace{0.17em}}8{x}^{2}-12xy+17{y}^{2}=20\text{\hspace{0.17em}}$ in the $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ system without an $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ term.

First, we find $\text{\hspace{0.17em}}\mathrm{cot}\left(2\theta \right).\text{\hspace{0.17em}}$ See [link] .

$\mathrm{cot}\left(2\theta \right)=\frac{3}{4}=\frac{\text{adjacent}}{\text{opposite}}$

So the hypotenuse is

$\begin{array}{r}\hfill {3}^{2}+{4}^{2}={h}^{2}\\ \hfill 9+16={h}^{2}\\ \hfill 25={h}^{2}\\ \hfill h=5\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$

Next, we find and

Substitute the values of and into and

and

Substitute the expressions for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ into in the given equation, and then simplify.

Write the equations with $\text{\hspace{0.17em}}{x}^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{\prime }\text{\hspace{0.17em}}$ in the standard form with respect to the new coordinate system.

$\frac{{{x}^{\prime }}^{2}}{4}+\frac{{{y}^{\prime }}^{2}}{1}=1$

[link] shows the graph of the ellipse.

#### Questions & Answers

Need help solving this problem (2/7)^-2
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
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12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
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Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
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Commplementary angles
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Nharnhar
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a perfect square v²+2v+_