I am given Focus (0,0) Directrix Y=3 eccentricity = 2
My rough sketch -
D2 = 2D1
$\sqrt{x^{2}+y^{2}}=2\ \sqrt{\left(y-3\right)^{2}}$
and simplifies to
$x^{2}-3y^{2}+24y=36$
The correct answer (from text book) shows as $x^{2}-3y^{2}+16y=16$
By graphing both the original equation and the result after manipulation, I see the result is consistent. Meaning my original set up was wrong. I am helping a precalc student with a series of these, and all others (6 with various combination of X or Y directrix) worked to match the book answer.
Let me confirm your result by writing $$x^2-3y^2+24y-36~~~~(1)$$ in standard form $$\frac{(y-4)^2}{4}-\frac{x^2}{12}=1,$$ which is vertical hyperbola. its excentricity is $$e=\sqrt{\frac{4+12}{4}}=2$$ its transverse axis is $x=0$ and the conjugate axix is $y=4.$ The directrix is $ y-4=\pm 2/2 \implies y=5,3$ its focui are give by $y-4=\pm 2.2,$ $x=0$ so the foci are given by $x=0$ where $y=0,8$. The foci are $(0,0),(0,8)$.
So one directrix of the hyperbola is $y=3$ and its ecentricity is 2 one focus is $(0,0).$ Finally, your answer (1) is correct. See the Fig.