Question 426088
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Hi
standard form of  
9x^2- 16y^2 - 18x + 96y - 279=0    |Note use of ^ (uppercase 6)
 9[(x-1)^2 -1] -16(y-3)^2]-9] - 279 = 0
 9(x-1)^2 - 9 -16(y-3)^2 + 144 - 279 = 0
 9(x-1)^2-16(y-3)^2 = 144
 {{{(x-1)^2/16 - (y-3)^2/9 = 1}}}