SOLUTION: Find the foci of the graph. Then draw the graph. x^2/16 - y^2/36 = 1

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the foci of the graph. Then draw the graph. x^2/16 - y^2/36 = 1      Log On


   

Question 39227: Find the foci of the graph. Then draw the graph.
x^2/16 - y^2/36 = 1
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
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..

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