Question

Question #27455

Answer 1

The answer is: `(x-8/5)^2/(144/25)+y^2/(144/45)=1`.

Remembering that:

`x=rcostheta`
`y=rsintheta`

we can say: `x^2+y^2=r^2`.

Our function becomes:

`r=4/(3-2costheta)rArrr(3-2costheta)=4rArr`

`3r-2rcostheta=4rArr3sqrt(x^2+y^2)-2x=4rArr`

`3sqrt(x^2+y^2)=2x+4rArr` (with `x>=-2`)

`9x^2+9y^2=4x^2+16x+16rArr`

`5x^2-16x+9y^2=16rArr`

`5(x^2-16/5x)+9y^2=16rArr`

`5(x^2-16/5x+64/25-64/25)+9y^2=16rArr`

`5(x^2-16/5x+64/25)-64/5+9y^2=16rArr`

`5(x-8/5)^2+9y^2=16+64/5rArr`

`5(x-8/5)^2+9y^2=144/5rArr`

`(x-8/5)^2/(144/5*1/5)+y^2/(144/5*1/9)=1rArr`

`(x-8/5)^2/(144/25)+y^2/(144/45)=1`

That is an ellipse with center in `C(8/5,0)` and semi-axis `a=12/5` and `b=12/sqrt45=12/(3sqrt5)=4sqrt5/5`.