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I need to find the polar equation of an ellipse, with one of its foci at the pole (origin), with a horizontal major axis of $10$ units and a vertical minor axis of $6$ units.

So first off, the distance from the center to a vertex $a$ is half of the major axis, i.e. $a = 5$. Also, if $b = 3$ is half the minor axis, and $c$ is the distance from the center to a focus, we know $a^2 = b^2 + c^2$, so $c = 4.$ Therefore, the eccentricity is $$e = \frac{c}{a} = \frac{4}{5}.$$

Now, the general (polar) form for an ellipse with a horizontal major axis, with the left focus as the pole, is $$r(\theta) = \frac{ep}{1-e\cos \theta} = \frac{4p/5}{1-(4\cos \theta )/5}.$$ where $p$ is the distance from the directrix of the ellipse to the pole.

Now all that remains is to find $p$. We know that the ratio of the distance between a point and the focus to the distance between a point to the directrix is the eccentricity. If we take the point $(r(\pi), \pi)$, then the distance from the focus to the point is $r(\pi),$ and the distance from the point to the directrix is $p - r(\pi).$ Therefore $$\frac{r(\pi)}{p- r(\pi)} = \frac{4}{5}.$$

Also, we know from our general equation that $$r(\pi) = \frac{4p/5}{1-(4\cos \pi)/5} = \frac{4p/5}{1+4/5} = \frac{4p/5}{6/5} = 2p/3.$$

Plugging this into our earlier equation gives $$\frac{2p/3}{p-2p/3} = \frac{4}{5}$$ $$2 = \frac{4}{5},$$ which is absurd. So, what have I done wrong here?

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It should be $ 1+4/5=9/5$ but not $6/5$ in the denominator, so your equation gives $$ r(\pi)=\frac{4/5p}{1+4/5}=\frac{4p}{9} $$ and later it comes to $4/5=4/5$.

To get $p$, you need another approach. You know the $c=4$ and $a=5$ already. So $r(\pi)=a-c=1$. Remeber we just got $r(\pi)=4p/9$. So $p=9/4$.

p.s. In fact, the expression that $\rho=ep/(1-e\cos(\theta))$ automatically satisfies that $\rho(\pi)/(p-\rho(\pi))=e$ and $\rho(0)/(p+\rho(0))=e$, no matter what value $e$, $p$ takes. So you can never solve $e$ or $p$ from the approach you attempted.

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