12.07 Converting between polar and rectangular equations

It is often useful to convert equations into different forms. Let's look at how to convert a rectangular equation into polar form first.

Worked examples

Question 1

Convert the rectangular equation $5x+y=2$5x+y=2 into a polar equation.

Think: The rectangular point $\left(x,y\right)$(x,y) corresponds with the polar point $\left(r,\theta\right)$(r,θ) where $x=r\cos\theta$x=rcosθ and $y=r\sin\theta$y=rsinθ. By substituting this information into the original rectangular equation we can convert it into polar form.

Do: Substituting the information from above we get:

$5x+y$5x+y	$=$=	$2$2	original rectangular equation
$5r\cos\theta+r\sin\theta$5rcosθ+rsinθ	$=$=	$2$2	substitution
$r\left(5\cos\theta+\sin\theta\right)$r(5cosθ+sinθ)	$=$=	$2$2	factor out $r$r as the GCF
$\frac{r\left(5\cos\theta+\sin\theta\right)}{5\cos\theta+\sin\theta}$r(5cosθ+sinθ)5cosθ+sinθ​	$=$=	$\frac{2}{5\cos\theta+\sin\theta}$25cosθ+sinθ​	solve for $r$r by dividing
$r$r	$=$=	$\frac{2}{5\cos\theta+\sin\theta}$25cosθ+sinθ​	simplify the left hand side

Reflect: Thus, the polar form of $5x+y=2$5x+y=2 is $r=\frac{2}{5\cos\theta+\sin\theta}$r=25cosθ+sinθ​.

 

Now let's try converting an equation in polar form into rectangular form.

Question 2

Convert the polar equation $r=8\cos\theta-4\sin\theta$r=8cosθ−4sinθ into a rectangular equation. State the equation with all terms on one side.

Think: Recall that the rectangular point $\left(x,y\right)$(x,y) corresponds with the polar point $\left(r,\theta\right)$(r,θ) where $x=r\cos\theta$x=rcosθ and $y=r\sin\theta$y=rsinθ and $x^2+y^2=r^2$x2+y2=r2.

Do: We need to make the original equation $r=8\cos\theta-4\sin\theta$r=8cosθ−4sinθ look more like $x^2+y^2=r^2$x2+y2=r2. We can start by creating an $r^2$r2 term on the left hand side.

$r$r	$=$=	$8\cos\theta-4\sin\theta$8cosθ−4sinθ	original polar equation
$r^2$r2	$=$=	$r\left(8\cos\theta-4\sin\theta\right)$r(8cosθ−4sinθ)	multiply both sides by $r$r
$r^2$r2	$=$=	$8r\cos\theta-4r\sin\theta$8rcosθ−4rsinθ	distribute $r$r
$r^2$r2	$=$=	$8x-4y$8x−4y	substitute $x=r\cos\theta$x=rcosθ and $y=r\sin\theta$y=rsinθ
$x^2+y^2$x2+y2	$=$=	$8x-4y$8x−4y	substitute $x^2+y^2=r^2$x2+y2=r2
$x^2-8x+y^2+4y$x2−8x+y2+4y	$=$=	$0$0	move all terms to one side

Reflect: Thus the rectangular form of $r=8\cos\theta-4\sin\theta$r=8cosθ−4sinθ is $x^2-8x+y^2+4y=0$x2−8x+y2+4y=0.

 

Practice questions

Question 3

Question 4

Question 5