How do you convert #r = 6/(3 - 4cos(theta))# into cartesian form?

Question

Brooklyn Anderson · Accepted Answer

#7x^2-9y^2+48x+36=0#

We know that the cartesian coordinate #(x,y)# of a point is related with its polar coordinate #(r,theta)# as follows:

#x=rcostheta and y=rsintheta->r=sqrt(x^2+y^2)#

The given equation

#r=6/(3-4costheta)#

#=>3r-4rcostheta=6#

#=>3sqrt(x^2+y^2)-4x=6#

#=>(3sqrt(x^2+y^2))^2=(6+4x)^2#

#=>9x^2+9y^2=36+48x+16x^2#

#=>16x^2-9x^2+48x-9y^2+36=0#

#=>7x^2-9y^2+48x+36=0#

This is the cartesian form of the given polar equation.

graph{7x^2-9y^2+48x+36=0}

The source for what follows is A. S. Adikesavan.

For information, the polar equation

#r = d/(sqrt(a^2 + b^2) + c(a cos theta + b sin theta))#

represents

( parabola ellipse hyperbola) according as

( abs c =1 abs c < 1 abs c > 1).

For d = 2, a = 2, b = 2, c = 2 giving

#r =`1/(1 + 2(cos theta + sin theta))# that represents a hyperbola.

graph{ ((x^2+y^2)^0.5-1 +2(x + y))( -(x^2+y^2)^0.5-1 +2(x + y)) = 0[-2 2 -2 2]}

Changing c to 1, in the above assignment, a parabola is traced.

graph{ (x^2+y^2)^0.5-1 +(x + y) = 0[-2 2 -2 2]}

Changing c to 0.5, in the above assignment, an ellipse is traced.

graph{ (x^2+y^2)^0.5-1 +0.5(x + y) = 0[-4 2 -4 2]}

James Adkins · Answer

#7x^2-9y^2+48x+36=0#

Use #(x, y) = r (cos theta, sin theta) and r=sqrt(x^2+y^2)#,

Here, cross-multiplying and rearranging,

#3r=3sqrt(x^2+y^2)=6+4r cos theta=6+4x#. Squaring,

#9(x^2+y^2)=(6+4x)^2=36+48x+16x^2#. Rearranging,

#7x^2-9y^2+48x+36=0#

This represents a hyperbola with eccentricity# e = 4/3#.

Jordan Armstrong · Answer

undefined To convert $ r = \frac{6}{3 - 4\cos(\theta)} $ into Cartesian form, follow these steps:

1. Replace $ r $ with its Cartesian form equivalent using the conversion formulas:
   $$ r = \sqrt{x^2 + y^2} $$

2. Rewrite the given polar equation with its Cartesian form equivalent:
   $$ \sqrt{x^2 + y^2} = \frac{6}{3 - 4\cos(\theta)} $$

3. Square both sides of the equation to eliminate the square root:
   $$ x^2 + y^2 = \frac{36}{(3 - 4\cos(\theta))^2} $$

4. Use the trigonometric identity $ \cos^2(\theta) + \sin^2(\theta) = 1 $ to express $ \cos(\theta) $ in terms of $ x $ and $ y $:
   $$ \cos(\theta) = \frac{x}{r} $$

5. Substitute $ \frac{x}{r} $ for $ \cos(\theta) $ in the equation:
   $$ x^2 + y^2 = \frac{36}{\left(3 - 4\frac{x}{r}\right)^2} $$

6. Substitute $ r = \sqrt{x^2 + y^2} $ into the equation:
   $$ x^2 + y^2 = \frac{36}{\left(3 - 4\frac{x}{\sqrt{x^2 + y^2}}\right)^2} $$

7. Simplify the equation and express it in terms of $ x $ and $ y $ only.

By following these steps, you can convert the polar equation $ r = \frac{6}{3 - 4\cos(\theta)} $ into Cartesian form.