Question

If (x-4)^2 + 4(y-3)^2 = 16, find the sum of distances from any fixed point on the curve of two foci

Answer 1

I believe you divide both sides by 16

Answer 2

So this?
\[\frac{ (x-4)^2 }{ 16 } + \frac{ (y-3)^2 }{ 4 } = 1\]

Answer 3

yep, then you find the horizontal and vertical axis, (a^2=16), (b^2=4) ---->

Answer 4

This is the step that I get stuck on.

Answer 5

ahh i get it now

Answer 6

16 is the major radius/axis, the square root is 4 ( to the center), so you multiply it by 2 and you get 8

Answer 7

@jojokiw3

Answer 8

do you kind of get it now?