Trig

(x+3)^2/4^2 + (y-2)^2/5^2
The location of the Foci are at:
a.(-3,-1) and (-3,-5)
b. (-3,2) and (-3,5)
c. (-3,-3) and (-3,7)
d. (-3,1) and (-3,-5)
I get (-3,2) and (-3,5) Am I correct?

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asked by Don
  1. Why did you label this "trig" ?

    Your expression should be an equation if it is to represent either an ellipse or a hyperbola.
    Should it be
    ....... = 1 ?

    secondly, you have "bouncing exponents"
    What is (x+3)^2/4^2 supposed to say?
    the way you typed it, it would simplify to (x+3)

    check and correct your typing.

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    posted by Reiny

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