Convert each equation to its rectangular form.r=−4cosθr=-4\cos\thetar=−4cosθ

( x + 2 ) 2 + y 2 = 4 (x+2)^2+y^2=4 ( x + 2 ) 2 + y 2 = 4

Convert each equation to its rectangular form. r = − 4 cos θ r=-4\cos\theta r = − 4 cos θ

In this problem, we're asked to convert the equation into its rectangular form, and the equation that we're given is r equals negative 4 times the cosine of theta. So let's go ahead and get started with our steps here, starting by trying to get r cosine theta, r sine theta, or r squared. Now here, since I have an r on this left side and I have a cosine theta on that right side, it seems like multiplying both sides by r may be a good choice here. So let's go ahead and do that, multiplying both sides by r. Now on the left that leaves me with r squared and that's equal to negative 4 times r cosine theta. So here I now have an r squared and an r cosine theta just like I wanted. So I can move on to step number 2 and replace that r cosine theta and r squared using these formulas that I know. So I know that r squared is going to end up giving me x squared plus y squared and that's equal to negative 4x. Now from here we've replaced that and we want to rewrite our equation in its standard form. So how are we going to do that here? We have an x and we have a y, and I'm not really sure what to solve for. But looking at our strategies here, I know that if I have x squared plus y squared is equal to some number times x or some number times y, I should go ahead and complete the square. So let's go ahead and do that here. I know that I have this negative 4x and I also have this x squared. So I want my x terms to be on the same side and I'm kind of going to ignore this y squared for a little bit. So if I add 4x to both sides it will cancel on that right side and I have x squared plus 4x and knowing that I'm going to have to complete the square. I'm just going to leave a blank right there and then that y squared is just off to the side and all of this is equal to 0. Now in order to complete the square remember that we want to get this down to a perfect square. And the specific perfect square that I'm looking for here is x+2squared because I have this 4x term. Now what do I need to add to this x squared+ 4 x in order to make this simplify to this perfect square? Well, I know that if I add 4, that will end up giving me that perfect square. But if I add 4 on that left side, I also need to add it on the right side. So I go ahead and do that. Now I can pull all of this stuff down, leaving me with that plus y squared and equals 4. Now I have x plus 2 squared plus y squared equals 4. And you might recognize this as being the equation of a circle in its standard form. So this is my equation in its rectangular form. Let me know if you have any questions here and thanks for watching.

Convert each equation to its rectangular form. r = − 4 cos ⁡ θ r=-4\cos\theta r = − 4 cos θ

( x + 2 ) 2 + y 2 = 4 (x+2)^2+y^2=4 ( x + 2 ) 2 + y 2 = 4

x 2 + y 2 = 4 x^2+y^2=4 x 2 + y 2 = 4

( x + 2 ) 2 + y 2 = 2 (x+2)^2+y^2=2 ( x + 2 ) 2 + y 2 = 2

15. Polar Equations

Convert Equations Between Polar and Rectangular Forms

Precalculus

15. Polar Equations

Convert Equations Between Polar and Rectangular Forms

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Multiple Choice

Convert each equation to its rectangular form.
$r=-4\cos\theta$

$r=-4x$

$x^2+y^2=4$

$(x+2)^2+y^2=2$

$(x+2)^2+y^2=4$

Verified step by step guidance

Start with the polar equation r = -4cos(θ). The goal is to convert this into a rectangular form, which involves x and y coordinates.

Recall the polar to rectangular coordinate conversions: x = r * cos(θ) and y = r * sin(θ). Also, r^2 = x^2 + y^2.

Substitute r = -4cos(θ) into the conversion formula for x: x = r * cos(θ) = -4cos(θ) * cos(θ) = -4cos^2(θ).

Use the identity cos(θ) = x/r to express cos^2(θ) in terms of x and r: cos^2(θ) = (x/r)^2.

Substitute cos^2(θ) = (x/r)^2 into the equation x = -4cos^2(θ) to get x = -4(x/r)^2. Simplify and rearrange to find the rectangular form: (x + 2)^2 + y^2 = 4.

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