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he graph of \(r = \frac{4}{2 - \cos \theta}\)

forms a closed curve. The area of the region inside the curve can be expressed in the form \(k \pi\). What is \(k^2\)?

 Dec 4, 2018
 #1
avatar+100439 
+2

Rekt.....let's put this in Cartesian form....maybe it will give us a clue as to what the graph might look like

 

√(x^2+ y^2)  =                4

                             ______________

                              2√(x^2 + y^2) -  x

                             _______________

                                  √(x^2 + y^2)

 

 

√ (x^2 + y^2)  =   4 √(x^2 + y^2)

                          _______________

                            2√(x^2 + y^2) - x

 

 

1           =               4

                         ______________

                        2√ (x^2 + y^2) - x 

 

 

2 √(x^2 + y^2) - x =  4

 

√(x^2 + y^2) - x/2 = 2

 

√(x^2 + y^2) =   2 + x/2     square both sides

 

x^2 + y^2 =   4 + 2x + x^2/4        multiply through by 4   

 

4x^2 + 4y^2 = 16 + 8x + x^2    simplify

 

3x^2 - 8x + 4y^2 =16          complete the square on x

 

3(x^2 - (8/3)x +  16/9   -16/9 ) + 4y^2 = 16

 

3(x - 4/3)^2 + 4 y^2  =  16 + 48/9

 

3(x - 4/3)^2  + 4y^2  =  64/3            multiply both sides by 3/64

 

(9/64) (x - 4/3)^2 + (3/16)y^2  =  1

 

(x - 4/3)^2             y^2

________   +      _____  =       1

   64/9                  16/3

 

 

This is an ellipse centered at (4/3, 0)

a = √[64/ 9] =  8/3

b = √[16/3] =  4/√3

 

The area is given by

 

pi * a * b   =

 

pi * 8/3 * 4/√3    =

 

pi * 32√3 / 9

 

k =  32√3 / 9

 

k^2 =    1024 / 27

 

 

cool cool cool

 Dec 4, 2018

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