Question

Convert r =-72/12+6 sin theta to Cartesian form.

Answer 1

The answer is b.
`4x^2+3y^2-24y-144=0`

Explanation:

The equation is:

`r=(-72)/(12+6\sin(\theta))`

Cross multiply;:

`r(12+6\sin(\theta)=-72`

Expand;:

`12r+6r\sin(\theta)=-72`

Divide through by 6;:

`2r+r\sin(\theta)=-12`

`2r=-12-r\sin(\theta)`

Substitute;:

`y=r\sin(\theta), r=√(x^2+y^2)`

`2√(x^2+y^2)=-12-y`

Square both sides;:

`(2√(x^2+y^2))^2=(-12-y)^2`

`4(x^2+y^2)=144+24y+y^2`

`4x^2+4y^2=144+24y+y^2`

Simplify;:

`4x^2+3y^2-24y-144=0`

Answer 2

For converting polar to cartesian form we know

`r = √(x^2 + y^2)`

`x = rcos \theta , y = rsin \theta`

Given equation is


`r = (-72)/(12) + 6sin \theta`

We can simplify it as

`r = -6 + 6sin \theta`

Now we can write it as

`r^2 = -6r + 6rsin \theta`

Now we can use

`r^2 = x^ 2+ y^2 , rsin \theta = y , r = √(x^2 + y^2)`
So equation we can write it as

`x^2 + y^2 = -6 √(x^2 + y^2) + 6y`

`x^2 + y^2 - 6y = -6 √(x^2 + y^2)`
On squaring both sides

`(x^2 + y^2 - 6y)^2 = (-6 √(x^2 + y^2))^2`

`x^4 + y^4 + 36y^2 +2x^2y^2 -12y^3 -12x^2y = 36(x^2 + y^2)`

`x^4 + y^4 +36y^2 + 2x^2y^2 -12y^3 -12x^2y = 36x^2 + 36y^2`

`x^4 + y^4 + 36y^2 +2x^2y^2 -12y^3-12x^2y - 36x^2 - 36y^2 = 0`

`x^4 + y^4 -12y^3 + 2x^2y^2 - 12x^2y - 36x^2 = 0`