SOLUTION: Write the equation of a hyperbola from the given information. Graph the equation. Place the center of the hyperbola at the orgin of the coordinate plane. One focus is located at (0

Question 39217: Write the equation of a hyperbola from the given information. Graph the equation. Place the center of the hyperbola at the orgin of the coordinate plane. One focus is located at (0,6); one vertex at (0,-(square root) 7
Found 2 solutions by Nate, venugopalramana:
Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
c(0,0)
f(0,6) this focus means that the hyperbola has a vertical transverse axis
v(0,-sqrt(7))
a = +-sqrt(7)
c = 6
c^2 = a^2 + b^2
6^2 = (sqrt(7))^2 + b^2
36 = 7 + b^2
29 = b^2
+-sqrt(29) = b

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Write the equation of a hyperbola from the given information. Graph the equation. Place the center of the hyperbola at the orgin of the coordinate plane. One focus is located at (0,6); one vertex at (0,-(square root) 7
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THE EQN. OF A HYPERBOLA IN STD.FORM IS
(X-H)^2/A^2 - (Y-K)^2/B^2=-1�.
WHERE
(H,K) IS CENTRE�HERE WE HAVE CENTRE IS (0,0).SO H=K=0
VERTICES ARE (0,B),(0,-B)...B=SQRT(7)
FOCI ARE (H,K+BE)AND (H,K-BE),THAT IS (0,BE),(0,-BE)....THEY ARE (0,6) AND (0,-6) SINCE CENTRE IS ORIGIN.HENCE BE=6.......E=6/SQRT(7)
E^2=36/7=(A^2+B^2)/B^2=(A^2+7)/7
A^2+7=36....A^2=29
HENCE EQN.IS
X^2/29-Y^2/7=-1...OR....Y^2/7-X^2/29=1

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