Proving Kepler's 2nd Law: Pluto's Area Sweep

[Trizz]

Homework Statement

I have to prove that A1=A2, or the fact that planets will sweep out equal areas in equal amounts of time. The planet I have to do this for is Pluto. Basically, I need help finding the area it "sweeps out" over any length of time of my choosing.

Homework Equations

major axis = 11.8 x 10^9 km
eccentricity = .25
minimum orbital velocity = 3.7 km/s

The Attempt at a Solution

So far, I've found the distance of the perihelion and the aphelion.

To do this I did: eccentricity = distance b/w foci / major axis

then i solved for the foci by doing: dis b/w foci = .25 * 11.8e9

that gave me the distance between the foci, so I was able to find out that the perihelion is 4425000000 km away and the aphelion is 7375000000 km.

Now do I just make a triangle for each with the legs being the distance away and the other being time traveled. It's hard to explain, but this picture shows it well:

http://outreach.atnf.csiro.au/education/senior/cosmicengine/images/cosmoimg/kepler2ndlaw.gif

So basically, how do I prove that those two areas equal with the info I now have

Any help would be appreciated.

 

[Trizz]

I believe this equation may help

Area=1/2 x (Distance to sun) x (Current Velocity) x (Time)


so would it be

(.5)(perihelion dis)(?)(anything)

then

(,5)(aphelion dis)(?)(anything)

I put "?"'s in the velocity spot because I don't know how to find the velocity

 

[Trizz]

I just noticed that it gives me Pluto's minimum orbital velocity at 3.7 km/s. So would plugging that into the formula above, the area one, give me the area of the aphelion (since it would be going slowest then)? Also, how would I determine the velocity of the perihelion.