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In an ellipse, I am given that the shortest distance from one focus to the circumference is $1.5$. Then the longest distance from that focus to the circumference of $4.5$. What is the area of the ellipse?

I'm pretty sure the semi-major axis would be $3$, but I do not know how to calculate the semi-minor axis now.

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    In a ellipse, there is no circumference. Do you mean that $1.5$ and $4.5$ are the lengths of the shortest and longest segment from a focus to the boundary? In such a case, you have $a=3$ and $c=\frac{3}{2}$, hence $b=\sqrt{a^2-c^2}=3\frac{\sqrt{3}}{2}$ and the area is $A = \pi a b = \frac{9\pi \sqrt{3}}{2}.$ – 2017-01-02
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    Yes, sorry I meant boundary. Apparently, the answer is 6.00pi. I still do not understand how to get that answer. – 2017-01-02
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    The original question was this: "If the closest distance from a planet to its host star is 1.50 AU and its farthest distance from its host star is 4.50 AU, what is the area that this planet sweeps out over the course of a full orbit (in AU2)?" – 2017-01-02

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Hint...if you already know that $a=3$ then you probably also have that the eccentricity $e=\frac 12$

In which case you can use $$b^2=a^2(1-e^2)$$ and the area is $\pi ab$