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
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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
{{{ graph( 600, 600, -10, 10, -10, 10,((1+x^2/29)*7)^0.5,-((1+x^2/29)*7)^0.5) }}}