# 8.4 Rotation of axes  (Page 3/8)

 Page 3 / 8

## 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.

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
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
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.
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
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_