# Thread: show that this is an ellipse.

1. ## show that this is an ellipse.

Given tha:

$x=4+4\cos \theta$

$y+3=5\sin \theta$

Show that the above equations represents an ellipse.

2. Originally Posted by razemsoft21
Given tha:

$x=4+4\cos \theta$

$y+3=5\sin \theta$

Show that the above equations represents an ellipse.
solving each equation for the trig functions gives.
$\frac{x-4}{4}=\cos(\theta)$ and
$\frac{y+3}{5}=\sin(\theta)$
And Remember that
$\sin^2(\theta)+\cos^2(\theta)=1$

3. Originally Posted by TheEmptySet
solving each equation for the trig functions gives.
$\frac{x-4}{4}=\cos(\theta)$ and
$\frac{y+3}{5}=\sin(\theta)$
And Remember that
$\sin^2(\theta)+\cos^2(\theta)=1$
$\frac {(x-4)^2}{4^2}+\frac {(y+3)^2}{5^2}=1$

Thats clear, but why this method works
all we do is finding $\sin^2 \theta+\cos^2 \theta$ wich is always = 1

and $\sin^2 \theta+\cos^2 \theta =r^2$ in polar coordinates
I think we do not porve it this way !!
what do you say ?

4. Originally Posted by razemsoft21
$\frac {(x-4)^2}{4^2}+\frac {(y+3)^2}{5^2}=1$

Thats clear, but why this method works
all we do is finding $\sin^2 \theta+\cos^2 \theta$ wich is always = 1

and $\sin^2 \theta+\cos^2 \theta =r^2$ in polar coordinates
I think we do not porve it this way !!
what do you say ?
I say you are mistaken.

correction ...

$\sin^2 \theta+\cos^2 \theta \ne r^2$

$\sin^2 \theta+\cos^2 \theta = 1$ always. It's an identity.

... also $x^2 + y^2 = r^2$