Find focus, directrix, and graph of a parabola

Question

Find vertex focus and directrix of parabola $$9x^2+16y^2-24xy-18x-101y+109=0.$$ Then sketch the graph.

My work. $3 x -4 Y^2 =18 x + 101 y - 109$. Here $Y=3x-4y$ and $X=18 x + 101 y - 109$ then I cannot understand what to do.

Answer 1

After taking a rotation of $\alpha=\arctan(3/4)$ around the origin, that is $$x=\frac{4u-3v}{5}\quad y=\frac{3u+4v}{5},$$ the parabola becomes $$25v^2-70v-75u+109=0,$$ or $$Y=(u-4/5)=\frac{(v-7/5)^2}{3}=\frac{1}{3}X^2.$$ Now recall that for parabola of the form $Y=AX^2$, the focus is the point $(0,\frac{1}{4A})$ and the directrix is the line $Y=-\frac{1}{4A}$. Hence, going back with the substitutions, we obtain that the focus is $$(X,Y)=\left(0,\frac{1}{4(1/3)}\right)\rightarrow (v,u)=\left(\frac{7}{5},\frac{1}{4(1/3)}+\frac{4}{5}\right)\rightarrow (x,y)=\left(\frac{2}{5},\frac{41}{20}\right).$$ Are you able to find the directrix?