# Polar equation

• Feb 14th 2007, 03:17 PM
mike1
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]

y[squared]=1-2x
x[squared]-x+y[squared]=1
x[squared]+y[squared]=1
x[squared]+x-y=1
• Feb 14th 2007, 04:04 PM
topsquark
Quote:

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]

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}

sqrt{x^2 + y^2} = 1

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

-Dan
• Feb 15th 2007, 09:10 AM
Soroban
Hello, Mike!

Quote:

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

. . .= .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: . . . . . . .= .1 - 2x

(Dan, you got a unit circle. .That equation would be: r = 1)
• Feb 15th 2007, 09:49 AM
topsquark
I didn't see the 1 + cos(t) part. :o

-Dan