Consider the equation below. $$ r=\frac{13}{4+2 \cos ( heta)} $$ (a) Find the eccentricity. $$ e=\square $$

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Consider the equation below. $$ r=\frac{13}{4+2 \cos (	heta)} $$ (a) Find the eccentricity. $$ e=\square $$

UpStudy AI · Accepted Answer

To find the eccentricity of the given polar equation $$ r=\frac{13}{4+2 \cos (\theta)} $$, we need to convert it into the standard form of a conic section in polar coordinates, which is given by:

$$ r = \frac{l}{1 + e \cos(\theta)} $$

where $$ l $$ is the semi-latus rectum and $$ e $$ is the eccentricity.

Let's compare the given equation with the standard form:

$$ r = \frac{13}{4+2 \cos (\theta)} $$

We can rewrite this as:

$$ r = \frac{13}{4(1 + \frac{1}{2} \cos (\theta))} $$

Now, we can see that the term inside the parentheses in the denominator is similar to the standard form, with $$ \frac{1}{2} $$ being the coefficient of $$ \cos(\theta) $$. This means that $$ e $$ (the eccentricity) is equal to $$ \frac{1}{2} $$, because the standard form has $$ e \cos(\theta) $$ in the denominator.

Therefore, the eccentricity $$ e $$ is:

$$ e = \frac{1}{2} $$

So, the answer is:

$$ e = \frac{1}{2} $$
$$ e = \frac{1}{2} $$

Consider the equation below.

(a) Find the eccentricity.

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Consider the equation below. (a) Find the eccentricity.