Question

Kurva parabola y=1/2x^2-4x-6 mempunyai titik fokus

Answer 1

Y = 1/2 x² - 4x - 6
1/2 x² - 4x = y - 6.....kalikan 2
x² -8x = 2y - 12
(x-4)² = 2y -12 + 16
(x-4)² = 2y + 4
(x-4)² = 2(y+2) --> (x-a)² = 4p(y-b)  --> F { (a, b+p)}
a = 4 
4p = 2 ---> p = 1/2
b = - 2

F= { a, (b+p)} ---> F( 2, (-2+1/2)} = (2, - 3/2)