Answer
$${\text{hyperbola}}$$
Work Step by Step
$$\eqalign{
& 9{x^2} - 18x - 4{y^2} - 16y - 43 = 0 \cr
& {\text{Add 43 to both sides}} \cr
& 9{x^2} - 18x - 4{y^2} - 16y = 43 \cr
& \left( {9{x^2} - 18x} \right) - \left( {4{y^2} - 16y} \right) = 43 \cr
& 9\left( {{x^2} - 2x} \right) - 4\left( {{y^2} - 4y} \right) = 43 \cr
& {\text{Complete the square}} \cr
& 9\left( {{x^2} - 2x + 1} \right) - 4\left( {{y^2} - 4y + 4} \right) = 43 + 9\left( 1 \right) - 4\left( 4 \right) \cr
& 9{\left( {x - 1} \right)^2} - 4{\left( {y - 2} \right)^2} = 36 \cr
& {\text{Divide both sides by 36}} \cr
& \frac{{{{\left( {x - 1} \right)}^2}}}{4} - \frac{{{{\left( {y - 2} \right)}^2}}}{9} = 1 \cr
& {\text{The equation is written in the form }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1\,\,\, \cr
& {\text{Therefore,}} \cr
& {\text{The graph of the equation is a hyperbola}} \cr} $$
