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

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- February 14th 2007, 03:17 PMmike1Polar 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 - February 14th 2007, 04:04 PMtopsquark
- February 15th 2007, 09:10 AMSoroban
Hello, Mike!

Quote:

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*) - February 15th 2007, 09:49 AMtopsquark
I didn't see the 1 + cos(t) part. :o

-Dan