<PRE><B>Question 39227</B>
X^2/16 - Y^2/36 =1

THE EQN. OF A HYPERBOLA.STD.EQN.IS.
(X-H)^2/A^2 - (Y-K)^2/B^2=1..WHERE
(H,K) IS CENTRE.....(0,0) HERE
A=4 AND B=6
ECCENTRICITY =E =SQRT[(A^2+B^2)/(A^2)]=SQRT[(16+36)/16]=SQRT(13)/2

A*E=4*SQRT(13)/2=2SQRT(13)
FOCI ARE (H+-AE,K)......(+2SQRT(13),0)
AND ......(-2SQRT(13),0)

GRAPH IS GIVEN BELOW..
{{{graph( 500, 500, -10, 10, -10, 10,
(2.25*(x^2-16))^0.5,-(2.25*(x^2-16))^0.5)}}}
 --------------------------------------------------------</PRE>