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

Algebra ->  Quadratic-relations-and-conic-sections -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
y%5E2=+5x%5E2%2B25
or y%5E2-5x%5E2=+25
or %28y%5E2-5x%5E2%29%2F25=25%2F25%29
or y%5E2%2F25+-+x%5E2%2F5+=+1
or y%5E2%2F5%5E2+-+x%5E2%2F%28sqrt%285%29%29%5E2+=+1

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
y%5E2%2Fa%5E2+-+x%5E2%2Fb%5E2+=+1
where '2a' and '2b' are the lengths of transverse and conjugate axes respectively we have
a = 5, b = sqrt%285%29

Now, eccentricity is e=sqrt%281+%2B+b%5E2%2Fa%5E2%29
or e=sqrt%281+%2B+5%2F5%5E2%29
or e=sqrt%281+%2B+1%2F5%29
or e=sqrt%286%2F5%29

The foci are at (0,ae) and (0,-ae) i.e. (0,5sqrt%286%2F5%29) and (0,-5sqrt%286%2F5%29) i.e. (0,sqrt%2830%29) and (0,-sqrt%2830%29).


Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
y%5E2+=+5x%5E2+%2B+25
y%5E2+-+5x%5E2+=+25
%28y%5E2%2F25%29+-+%285x%5E2%2F25%29+=+25%2F25
%28y%5E2%2F25%29+-+%28x%5E2%2F5%29+=+1
This is a hyperbola.
Transverse axis: vertical
Center: (0,0)
a = 5
b = sqrt%285%29
Vertex:(0,5) and (0,-5)
a%5E2+%2B+b%5E2+=+c%5E2
5%5E2+%2B+%28sqrt%285%29%29%5E2+=+c%5E2
25+%2B+5+=+c%5E2
sqrt%2830%29+=+c
Foci: (0,sqrt%2830%29) and (0,-sqrt%2830%29)
Asymptote:
slope: (+-a/b) = +-5%2Fsqrt%285%29 = +-%285%2Asqrt%285%29%29%2F%28sqrt%285%29%2Asqrt%285%29%29 = +-sqrt%285%29
center (again): (0,0)
y+-+y1+=+m%28x+-+x1%29
y+-+0+=+sqrt%285%29%28x+-+0%29 and y+-+0+=+-sqrt%285%29%28x+-+0%29
y+=+sqrt%285%29x and y+=+-sqrt%285%29x