3(y-3)=(x+6)squared how do you find the vertex and foci,directrix

Answer 1

3(Y-3) = (X+6)^2,
3Y-9 = X^2+12X+36,
3Y = X^2+12X+36+9,
3Y = X^2+12X+45,
Y = X^2/3+4X+15,

h = -b/2a = -4/(2/3) = -12/2 = -6.

k=Yv = (-6)^2/3+4*(-6)+15 = 12-24+15 = 3.

V(h,k)= V(-6,3).

F(-6,Y2)

V(-6,3)

D(-6,Y1)

3-Y1 = 1/4a = 1/(4/3) = 3/4,
-Y1 = 3/4 -12/4 = -9/4,
Y1 = 9/4.

Y2-3 = 1/4a = 1/(4/3) = 3/4,
Y2 = 3/4+12/4 = 15/4.