SOLUTION: "Graph the equation, then choose the correct coordinates of the foci. Keep the graphs as they may be asked for at a later time. y^2= 5x^2+25." I keep trying to graph equations like

Question 43057: "Graph the equation, then choose the correct coordinates of the foci. Keep the graphs as they may be asked for at a later time. y^2= 5x^2+25." I keep trying to graph equations like this and find the foci, but the answers I get are never one of the answer choices. For this one I did:
y^2= 5x^2+25

y^2-5x^2= 25

(y^2-5x^2)/-5=25/-5
-1/5(y^2/-5-x^2)=-5/-5

y^2/25+x^2/5=1
y^2/(5)^2+x^2/(square root 5)^2=1
After that I had my x and y intercepts so I made my graph and tried to find the foci.
b^2=a^2-c^2
square root 5^2= 5^2-c^2
5= 25-c^2
5-25= -c^2
20= c^2
square root 20= c
I'd appreciate your help. These graphs have really been frustrating me. Please me know what I'm doing wrong. Thanx!
Found 2 solutions by psbhowmick, Nate:
Answer by psbhowmick(878) (Show Source): You can put this solution on YOUR website!

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As simple as that!
Don't make the algebra complex.
Divide both sides by 25 (to make the term on right side = 1).
This is a hyperbola and not an ellipse.
You derived a wrong equation resulting in wrong type of conic (you got ellipse).

Comparing with standard equation of hyperbola

where '2a' and '2b' are the lengths of transverse and conjugate axes respectively we have
a = 5, b =

Now, eccentricity is
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The foci are at (0,ae) and (0,-ae) i.e. (0,) and (0,) i.e. (0,) and (0,).

Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!

This is a hyperbola.
Transverse axis: vertical
Center: (0,0)
a = 5
b =
Vertex:(0,5) and (0,-5)

Foci: (0,) and (0,)
Asymptote:
slope: (+-a/b) = +- = +- = +-
center (again): (0,0)

and
and

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