Thread: Polar equation

I need help with this one:

Find an equation in x and y for the conic section with polar equation r=
1
___
1+cos[theta]

possible answers:
y[squared]=1-2x
x[squared]-x+y[squared]=1
x[squared]+y[squared]=1
x[squared]+x-y=1

Originally Posted by mike1

I need help with this one:

Find an equation in x and y for the conic section with polar equation r=
1
___
1+cos[theta]

possible answers:
y[squared]=1-2x
x[squared]-x+y[squared]=1
x[squared]+y[squared]=1
x[squared]+x-y=1

Well,
x = r*cos(t)
y = r*sin(t)

Thus x^2 + y^2 = r^2*cos^2(t) + r^2*sin^2(t) = r^2(cos^2(t) + sin^2(t)) = r^2

So r = sqrt{x^2 + y^2}

Thus your equation is
sqrt{x^2 + y^2} = 1

Squaring both sides gives,
x^2 + y^2 = 1, which is your third option.

-Dan

Hello, Mike!

Find an equation in x and y for the conic section with polar equation:
. . . . . . . . . 1
. . r .= . -----------
. . . . . . 1 + cos θ

Possible answers:

. . y² .= .1 - 2x . . x² - x + y² .= .1 . . x² + y² .= .1 . . x² + x - y .= .1

We have: .r(1 + cosθ) .= .1

. . . . . . . . .r + r·cosθ .= .1

. . . . . . . . . . . . . . .r .=] .1 - r·cosθ
. . . . . . . . . . ______
Convert: . . .√x² + y² .= .1 - x

Square: . . . . .x² + y² .= .1 - 2x + x²

Therefore: . . . . . . y² .= .1 - 2x


(Dan, you got a unit circle. .That equation would be: r = 1)

I didn't see the 1 + cos(t) part.

-Dan