Question 21538: Find the foci for the given hyperbola.
9y2-x2-36y-2x+26=0
y2=ysquared
x2=xsquared
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 9y^2-x^2-36y-2x+26=0
{(3Y)^2-2*3Y*6+6^2}-6^2-{X^2+2*X*1+1}-1+26=0
(3Y-6)^2-(X+1)^2-9=0
9(Y-2)^2-(X+1)^2=9...DIVIDING BY 9 THROUGH OUT
((Y-2)^2)/1-((X+1)^2)/9=1..COMPARING WITH THE STANDARD EQUATION
((Y-K)^2)/(B^2)-((X-H)^2)/(A^2)=1...WE GET...THE CENTRE (H,K)=(-1,2)
ECENTRICITY =E = SQUAREROOT{(A^2+B^2)/(B^2)}=SQ.RT{(9+1)/1}=SQ.RT(10)
HENCE FOCI ARE ...(H,K+B*E)AND(H,K-B*E)
{-1,(2+SQ.RT(10))} AND {-1,2-SQ.RT(10)}
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