SOLUTION: I was wondering if anyone could help me determine what type of figure is given, then graph. Show all work and label all key points (asymptotes, foci, vertices, the directrix, the c

Question 36550: I was wondering if anyone could help me determine what type of figure is given, then graph. Show all work and label all key points (asymptotes, foci, vertices, the directrix, the center) where applicable.
9x^2 � 96y = 16y^2 + 18x + 279
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
TIP
IF X^2 AND Y^2 HAVE DIFFERENT COEFFICIENTS WITH OPPOSITE SIGNS THEN IT
COULD BE HYPERBOLA
3.) 9x2 � 96y = 16y2 + 18x + 279
COMPLETE SQUARE...
{(3X)^2-2*3X*3+3^2}-3^2-{(4Y)^2+2*4Y*12+12^2}-12^2=279
(3X+3)^2-(4Y+12)^2=279+9+144=432
9(X+1)^2-16(Y+3)^2=432......NOW DIVIDE THROUGH OUT WITH 432 TO GET 1 ON THE RHS.
(X+1)^2/(432/9) -(Y+3)^2/(432/16)=1
(X+1)^2/48 - (Y+3)^2/27 =1
THIS THE EQN. OF A HYPERBOLA.STD.EQN.IS.
(X-H)^2/A^2 - (Y-K)^2/B^2=1..WHERE
(H,K) IS CENTRE.....(-1,-3) HERE
TRANSVERSE AXIS IS Y=K...Y=-3
LENGTH OF TRANSVERSE AXIS=2A..
...=2SQRT(48)
CONJUGATE AXIS IS X=H.......X=-1
LENGTH OF CONJUGATE AXIS = 2B
=2SQRT(27)
ECCENTRICITY=E=SQRT{(A^2+B^2)/A^2}
=SQRT{(48+27)/48}=SQRT(75/48)
A*E=SQRT(48)*SQRT(75/48)=SQRT(75)
FOCI ARE (H+-AE,K)......(-1+SQRT(75),-3)
AND ......(-1-SQRT(75),-3)
A/E=SQRT(48)/SQRT(75/48)=48/SQRT(75)
DIRECTRIX ARE X=H+-A/E....
X=-1+48/SQRT(75)...AND
X=-1-48/SQRT(75)
ASYMPTOTES ARE GIVEN BY
(X-H)^2/A^2 = (Y-K)^2/B^2
OR
(X-H)/A=+(Y-K)/B AND............(X+1)/SQRT(48) =(Y+3)/SQRT(27)
(X-H)/A=-(Y-K)/B.............(X+1)/SQRT(48) = -(Y+3)/SQRT(27)
GRAPH IS GIVEN BELOW..

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