Given the general form of a parabola x2+8x-24y+16=0, what is the x coordinate the focus? (round to 1 decimal places)

Question

Given the general form of a parabola x²+8x-24y+16=0, what is the x coordinate of the focus? (round to 1 decimal places)

Answer 1

y= (1/24)(x^2 +8x +16)

= (1/24)(x^2 + 8x + 16) + 16/24 -16/24

= (1/24)(x+4)^2

vertex is (-4,0)

focus has the same x coordinate, as it's an upward opening parabola

y'= 2x/24 +8/24 =0

x/12 + 1/3 = =

x/12 =-1/3

x = -12/3 = -4 = x coordinate of the vertex

y = (1/24)(x^2+8x + 16)

= (1/24)(16 -32 +16) = 0 = y coordinate of the vertex

24 =4p

p = 8

y coordinate of the focus = 8+0 = 8

focus is (-4,8)

directrix is y=-8

vertex midway between them (-4,0)