# Thread: Identifying graph of a polar equation.

1. ## 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?

2. What are e and p here??

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

Spoiler:

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

$\displaystyle 4r-3rcos\theta=3$

$\displaystyle 4r=3+3x$

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

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

3. 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?

4. e is the eccentricity and |p| is the distance between the focus and the directrix.

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
$\displaystyle r = \frac{ep}{1 - e\cos \theta}$

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

The denominator is $\displaystyle 1 - 0.75\cos \theta$, 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).

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 $\displaystyle \theta = \pi/2$, $\displaystyle \cos \pi/2 = 0$, so

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

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

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

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

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

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

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

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there is typo in the above

6. Fixed. Thanks.

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