Calculus (3rd Edition)

$a \ge - 44$
Write $3{x^2} + 2{y^2} - 16y + 12x = a$ $3\left( {{x^2} + 4x} \right) + 2\left( {{y^2} - 8y} \right) = a$ $3\left( {{x^2} + 4x + 4} \right) - 12 + 2\left( {{y^2} - 8y + 16} \right) - 32 = a$ $3{\left( {x + 2} \right)^2} + 2{\left( {y - 4} \right)^2} = a + 44$ Since the left-hand side is nonnegative, we must have $a + 44 \ge 0$ so that the conic has at least one point. Thus, $a \ge - 44$