SOLUTION: Rather confused about what's going on here. My book doesn't offer an explanation on what I should do with this type of set up. Write an equation for each ellipse. "foci at

Question 540637: Rather confused about what's going on here. My book doesn't offer an explanation on what I should do with this type of set up.
Write an equation for each ellipse.
"foci at (-3,-3), (7,-3); the point (2,-7) on ellipse"
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Write an equation for each ellipse.
"foci at (-3,-3), (7,-3); the point (2,-7) on ellipse
**
Given data shows equation is that of an ellipse with horizontal major axis of the standard form
(x-h)^2/a^2+(y-k)^2/b^2=1, a>b, (h,k) being the (x,y) coordinates of the center. (x-coordinates of foci changes while y-coordinates do not).
..
For given equation:
x-coordinate of center=(7+(-3))/2=4/2=2
y-coordinate of center=-3
center:(2,-3)
c=distance from center to focus on major axis=5
c^2=25
..
solving for a and b in standard form of equation for ellipse:
Sum of the distance from point (2,-7) on ellipse to each of the two focal points=2a
Using distance formula:√[(x1-x2)^2+(y1-y^2)^2]+√[(x1-x2)^2+(y1-y^2)^2]=2a
√[(2+3)^2+(-7+3)^2]+√[(2-7)^2+(-7+3)^2]=2a
√[(5)^2+(-4)^2]+√[(-5)^2+(-4)^2]=2a
√[25+16]+√[25+16]=2a
√[41]+√[41]=2√41=2a
a=√41
a^2=41
..
c^2=a^2-b^2
b^2=a^2-c^2=41-25=16
..
Equation of given ellipse:
(x-2)^2/41+(y+3)^2/16=1
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