Question

3x2 − 4y2 + 6x + 24y − 135 = 0

Answer 1

$3x^2 − 4y^2 + 6x + 24y − 135 = 0$

$3(x^2 +2x+1-1)−4(y^2 - 6y+9-9) − 135 = 0$

$3(x-1)^2−4(y - 3)^2 − 168 = 0$

$(x-1)^2/56-(y-3)^2/42=1$

Thus, the conic represented by the given equation is a hyperbola whose focus is at (1,3).