# 12.4 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.

if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
hi
John
hi
Grace
what sup friend
John
not much For functions, there are two conditions for a function to be the inverse function:   1--- g(f(x)) = x for all x in the domain of f     2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the  element that you started with, namely, x.
Grace
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
acha se dhek ke bata sin theta ke value
Ajay
sin theta ke ja gha sin square theta hoga
Ajay
I want to know trigonometry but I can't understand it anyone who can help
Yh
Idowu
which part of trig?
Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
hi
Ganapathi
hello
Brittany
Prove that 4sin50-3tan 50=1
f(x)= 1 x    f(x)=1x  is shifted down 4 units and to the right 3 units.
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
what are real numbers
I want to know partial fraction Decomposition.
classes of function in mathematics
divide y2_8y2+5y2/y2
wish i knew calculus to understand what's going on 🙂
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
thanks bro
Dashawn
maybe when i start calculus in a few months i won't be that lost 😎
Dashawn
what's the derivative of 4x^6
24x^5
James
10x
Axmed
24X^5
Taieb
secA+tanA=2√5,sinA=?
tan2a+tan2a=√3
Rahulkumar
classes of function
Yazidu
if sinx°=sin@, then @ is - ?
the value of tan15°•tan20°•tan70°•tan75° -
NAVJIT