diketahui persamaan parabola y^2 - 6y - 8x + 1 = 0. Tentukan: - titik fokus - sumbu simetri - persamaan direktris

[tex]y^2-6y-8x+1=0 \\ (y^2-6y)-8x+1=0 \\ (y^2-6y+9)-8x+1-9=0 \\ (y-3)^2-8x-8=0 \\ (y-3)^2=8x+8 \\ (y-3)^2=8(x+1) \\ $Pusat : $(-1,3) \\ \\ 4p=8 \\ p=2 \\ \\ $Fokus : $(-1+2,3)=(1,3) \\ \\ $Simetri : $y=3 \\ \\ $Diretriks : $x=-1-p \\ x=-1-2 \\ x=-3[/tex]

y^2-6y-8x+1=0
y^2-6y=8x-1
y^2-6y+9=8x-1+9
(y-3)^2=8x+8
(y-3)^2=8(x+1)
puncak = (-1,3) fokus (1,3)
sumbu simetri y=3
persamaan direktris x= -3