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The graph of an equation $$\sqrt{(x-3)^2 + (y+4)^2} + \sqrt{(x+5)^2 + (y-8)^2} = 20$$ is an ellipse. What is the distance between its foci?

Jan 14, 2019

#1
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This is an odd duck, indeed !!!!

It is a rotated ellipse

In the form

√ [ ( x - a)^2 + ( y - b)^2 ] +    √ [ (x - a ' )^2 + ( y - b ' )^2 ]  = C

The foci    are   located at   (a, b)  and ( a ' , b ' )

[ Notice how this is very similar  to finding the center of  any conic....!!! ]

So....the foci here are  ( 3, -4)  and (-5, 8)

So....the distance between these = √ [ ( -5 - 3)^2 + (-4 - 8)^2 ]   = √ [ 64 + 144]  = √ [ 208 ] = 4 √13 units

Jan 14, 2019