persmaan parabola y^2-6y+8×+1=0 memiliki koordinat titik fokus

jawab

y² - 6y = -8x - 1
(y - 3)² = -8x - 1 + 9
(y - 3)² = -8x + 8
(y - 3)³ = -8(x + 1) --> (y -b ) = 4p (x - a)
b= 3
4p  = -8 --> p = - 2
a = -1

Fokus (a+p, b)
F ((-1- 2), (3))
F (- 2, 3)