SOLUTION: find an equation of the hyberbola centered at the origin that has a focus at (square root of 5,0)

Question 38930: find an equation of the hyberbola centered at the origin that has a focus at (square root of 5,0)
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
find an equation of the hyberbola centered at the origin that has a focus at (square root of 5,0)
SEE THE FOLLOWING EXAMPLE FOR DETAILED ANALYSIS.
TAKE EQN.AS X^2/A^2-Y^2/B^2=1..............................I
GIVEN DATA IS INSUFFICIENT TO FIND AN UNIQUE HYPERBOLA..YOU CAN OFCOURSE GIVE EQN.OF ANY HYPERBOLA SATISFYING THESE CONDITIONS.
AE=5 HERE..TAKE ANY SET OF VALUES WITH E>1 SATISFYING THE ABOVE EQN.SAY A=2.5 AND E=2...FIND B FROM EQN.
E=SQRT(A^2+B^2)/A^2 ...SUBSTITUTE A AND B IN EQN.I TO GET THE ANSWER.BUT CHECK THE PROBLEM AGAIN
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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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