# Identifying graph of a polar equation.

• Jul 7th 2009, 09:05 PM
Wesker
Identifying graph of a polar equation.
Having a little trouble with where to go with this problem.

Identify the graph of the polar equation r = 3/(4-3cosθ)

3/4/(1-(3/4)cos θ,
e = -3/4
ep = 3/4
3p = 3/4
p = 1/4

Did i get these right? and if so how do i go from here to identify the graph?
I think θ= p/2 at r = 3/4/(1-3/4) = 3 so the graph runs through (p/2,3) right? What from here?
• Jul 7th 2009, 09:14 PM
malaygoel
What are e and p here??

To identify graph, try to convert in x-y cordinates.

Spoiler:

$r=\frac{3}{4-3cos\theta}$

$4r-3rcos\theta=3$

$4r=3+3x$

$16r^2=9+18x+9x^2$

$16x^2+16y^2=9+18x+9x^2$

• Jul 7th 2009, 09:24 PM
Wesker
r = 3 / 4 - 3 cos or ep/1 - e cos q
e would be 3 i think,(or 3/4 after you divide the denominator by 4) and Im not sure if got p right, but if the top is ep, correct me if im wrong but wouldn't it be 3p = 3/4? so p = 1/4?
• Jul 7th 2009, 11:15 PM
yeongil
e is the eccentricity and |p| is the distance between the focus and the directrix.

Quote:

Originally Posted by Wesker
Having a little trouble with where to go with this problem.

Identify the graph of the polar equation r = 3/(4-3cosθ)

The form we are interested in is
$r = \frac{ep}{1 - e\cos \theta}$

\begin{aligned}
r &= \frac{3}{4 - 3\cos \theta} \\
r &= \frac{0.75}{1 - 0.75\cos \theta}
\end{aligned}

The denominator is $1 - 0.75\cos \theta$, which means that it has a vertical directrix to the left of the pole.

e = 3/4 < 1, so the conic is an ellipse.

The numerator is ep = 3/4, so p = 1 (not 1/4).

Quote:

I think θ= p/2 at r = 3/4/(1-3/4) = 3 so the graph runs through (p/2,3) right? What from here?
No, at $\theta = \pi/2$, $\cos \pi/2 = 0$, so

$r = \frac{3}{4 - 3\cos (\pi/2)} = \frac{3}{4}$

From here, you might as well find r at the other quadrantal angles.
When $\theta = 0$,
$r = \frac{3}{4 - 3\cos 0} = ????$

When $\theta = \pi$,
$r = \frac{3}{4 - 3\cos \pi} = ????$

When $\theta = 3\pi/2$,
$r = \frac{3}{4 - 3\cos (3\pi/2)} = ????$

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• Jul 7th 2009, 11:20 PM
malaygoel
Quote:

Originally Posted by yeongil
When $r = 0$,
$r = \frac{3}{4 - 3\cos 0} = ????$

When $r = \pi$,
$r = \frac{3}{4 - 3\cos \pi} = ????$

When $r = 3\pi/2$,
$r = \frac{3}{4 - 3\cos (3\pi/2)} = ????$

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there is typo in the above
• Jul 7th 2009, 11:38 PM
yeongil
Fixed. Thanks.

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