SOLUTION: pls help. write hyperbola 25y^2-4x^2+100y+24x-36=0 in standard form center vertices foci and graph also 25y^2-x^2=25 I put center (0,0) vertices (0,1) (0,-1) What is fo

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: pls help. write hyperbola 25y^2-4x^2+100y+24x-36=0 in standard form center vertices foci and graph also 25y^2-x^2=25 I put center (0,0) vertices (0,1) (0,-1) What is fo      Log On


   

Question 451429: pls help.
write hyperbola 25y^2-4x^2+100y+24x-36=0 in
standard form
center
vertices
foci and graph
also 25y^2-x^2=25
I put center (0,0) vertices (0,1) (0,-1)
What is foci?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi, 

Note: Standard Form of an Equation of an Hyperbola opening up and down is:

  %28y-k%29%5E2%2Fb%5E2+-+%28x-h%29%5E2%2Fa%5E2+=+1 

where Pt(h,k) is a center  with vertices 'b' units up and down from center.

25y^2-4x^2+100y+24x-36=0 

 25[(y+2)^2 -4] -4[(x-3)^2 - 9] - 36 = 0

   25(y+2)^2 -100 - 4(x-3)^2 + 36 - 36 = 0

   25(y+2)^2 - 4(x-3)^2 = 100

%28y%2B2%29%5E2%2F4+-%28x-3%29%5E2%2F25+=1  C(3,-2) with vertices V(3,0) and V(3,-4)

Foci: c = sqrt(29)  foci(-3, -2-sqrt(29)) and (-3, -2+sqrt(29))



25y^2-x^2=25

 +y%5E2%2F1+-+x%5E2%2F25+=+1 Yes. C(0,0) vertices(0,1),(0,-1)... foci(0,ħsqrt(26))