Question 36550
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..
{{{ graph( 500, 500, -50, 50, -50, 50, -3+27*(((x+1)^2-48)/48)^0.5,-3-27*(((x+1)^2-48)/48)^0.5)}}} 

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