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

Find a new representation of the equation 2 x 2 x y + 2 y 2 30 = 0 after rotating through an angle of θ = 45° .

Find x and y , where x = x cos   θ y sin   θ and y = x sin   θ + y cos   θ .

Because θ = 45° ,

x = x cos ( 45° ) y sin ( 45° ) x = x ( 1 2 ) y ( 1 2 ) x = x y 2

and

y = x sin ( 45° ) + y cos ( 45° ) y = x ( 1 2 ) + y ( 1 2 ) y = x + y 2

Substitute x = x cos θ y sin θ and y = x sin   θ + y cos   θ into 2 x 2 x y + 2 y 2 30 = 0.

2 ( x y 2 ) 2 ( x y 2 ) ( x + y 2 ) + 2 ( x + y 2 ) 2 30 = 0

Simplify.

2 ( x y ) ( x y ) 2 ( x y ) ( x + y ) 2 + 2 ( x + y ) ( x + y ) 2 30 = 0 FOIL method             x 2 2 x y + y 2 ( x 2 y 2 ) 2 + x 2 + 2 x y + y 2 30 = 0 Combine like terms .                                                               2 x 2 + 2 y 2 ( x 2 y 2 ) 2 = 30 Combine like terms .                                                         2 ( 2 x 2 + 2 y 2 ( x 2 y 2 ) 2 ) = 2 ( 30 ) Multiply both sides by 2 .                                                                4 x 2 + 4 y 2 ( x 2 y 2 ) = 60 Simplify .                                                                   4 x 2 + 4 y 2 x 2 + y 2 = 60 Distribute .                                                                                      3 x 2 60 + 5 y 2 60 = 60 60 Set equal to 1 .

Write the equations with x and y in the standard form.

x 2 20 + y 2 12 = 1

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

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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 A x 2 + B x y + C y 2 + D x + E y + F = 0 into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the x and y coordinate system without the x y term, by rotating the axes by a measure of θ that satisfies

cot ( 2 θ ) = A C B

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

A x 2 + B x y + C y 2 + D x + E y + F = 0

where A , B , and C are not all zero. However, if B 0 , then we have an x y term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle θ where cot ( 2 θ ) = A C B .

  • If cot ( 2 θ ) > 0 , then 2 θ is in the first quadrant, and θ is between ( , 45° ) .
  • If cot ( 2 θ ) < 0 , then 2 θ is in the second quadrant, and θ is between ( 45° , 90° ) .
  • If A = C , then θ = 45° .

Given an equation for a conic in the x y system, rewrite the equation without the x y term in terms of x and y , where the x and y axes are rotations of the standard axes by θ degrees.

  1. Find cot ( 2 θ ) .
  2. Find sin   θ and cos   θ .
  3. Substitute sin   θ and cos   θ into x = x cos   θ y sin   θ and y = x sin   θ + y cos   θ .
  4. Substitute the expression for x and y into in the given equation, and then simplify.
  5. Write the equations with x and y 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 8 x 2 12 x y + 17 y 2 = 20 in the x y system without an x y term.

First, we find cot ( 2 θ ) . See [link] .

8 x 2 12 x y + 17 y 2 = 20 A = 8 , B = 12 and C = 17                   cot ( 2 θ ) = A C B = 8 17 12                   cot ( 2 θ ) = 9 12 = 3 4
cot ( 2 θ ) = 3 4 = adjacent opposite

So the hypotenuse is

3 2 + 4 2 = h 2 9 + 16 = h 2 25 = h 2 h = 5

Next, we find sin   θ and cos   θ .

sin   θ = 1 cos ( 2 θ ) 2 = 1 3 5 2 = 5 5 3 5 2 = 5 3 5 1 2 = 2 10 = 1 5 sin   θ = 1 5 cos   θ = 1 + cos ( 2 θ ) 2 = 1 + 3 5 2 = 5 5 + 3 5 2 = 5 + 3 5 1 2 = 8 10 = 4 5 cos   θ = 2 5

Substitute the values of sin   θ and cos   θ into x = x cos   θ y sin   θ and y = x sin   θ + y cos   θ .

x = x cos   θ y sin   θ x = x ( 2 5 ) y ( 1 5 ) x = 2 x y 5

and

y = x sin   θ + y cos   θ y = x ( 1 5 ) + y ( 2 5 ) y = x + 2 y 5

Substitute the expressions for x and y into in the given equation, and then simplify.

                                   8 ( 2 x y 5 ) 2 12 ( 2 x y 5 ) ( x + 2 y 5 ) + 17 ( x + 2 y 5 ) 2 = 20        8 ( ( 2 x y ) ( 2 x y ) 5 ) 12 ( ( 2 x y ) ( x + 2 y ) 5 ) + 17 ( ( x + 2 y ) ( x + 2 y ) 5 ) = 20          8 ( 4 x 2 4 x y + y 2 ) 12 ( 2 x 2 + 3 x y 2 y 2 ) + 17 ( x 2 + 4 x y + 4 y 2 ) = 100 32 x 2 32 x y + 8 y 2 24 x 2 36 x y + 24 y 2 + 17 x 2 + 68 x y + 68 y 2 = 100                                                                                                    25 x 2 + 100 y 2 = 100                                                                                                    25 100 x 2 + 100 100 y 2 = 100 100  

Write the equations with x and y in the standard form with respect to the new coordinate system.

x 2 4 + y 2 1 = 1

[link] shows the graph of the ellipse.

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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