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?

Guest Jan 14, 2019

#1**+1 **

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

CPhill Jan 14, 2019