# Thread: Ellipse..in cartesian & polar co-ordinates..

1. ## Ellipse..in cartesian & polar co-ordinates..

Given an ellipse C with foci f1=(0,0) and f2=(1,0) and the distances between an arbitrary point on C and its foci equals to 3.

I have to write down the equation in cartesian co-ordinates and polar co-ordinates..

For cartesian I've got [(x-0.5)^2] / (9/4) + (y^2) / (5/4) = 1
But I'm not sure if that's correct as I made calculations when it's meant to be as simple as writing it down
And I've no idea on the polar front...

Would be grateful of any help

2. Originally Posted by mathlete09
Given an ellipse C with foci f1=(0,0) and f2=(1,0) and the distances between an arbitrary point on C and its foci equals to 3.

I have to write down the equation in cartesian co-ordinates and polar co-ordinates..

For cartesian I've got [(x-0.5)^2] / (9/4) + (y^2) / (5/4) = 1
But I'm not sure if that's correct as I made calculations when it's meant to be as simple as writing it down
And I've no idea on the polar front...

Would be grateful of any help
The centre of the ellipse is at $\displaystyle \bigl(\tfrac12,0\bigr)$ and the semi-axes have length $\displaystyle a=3/2$ and $\displaystyle b = \sqrt{\bigl(\tfrac32\bigr)^2 - \bigl(\tfrac12\bigr)^2} = \sqrt2$, so the cartesian equation should be $\displaystyle \frac{\bigl(x-\frac12\bigr)^2}{9/4} + \frac{y^2}2 = 1$. For the polar equation, just substitute $\displaystyle r\cos\theta$ for x and $\displaystyle r\sin\theta$ for y.