Question

What are the vertex, focus, and directrix of the parabola with the equation x^2+y^2-8x+4y+4=0?

vertex (4, –3); focus (4, –4); directrix y = –2

vertex (–4, 3); focus (–4, –1); directrix y = 7 vertex (4, –3); focus (4, –7); directrix y = 1 vertex (–4, 3); focus (–4, 2); directrix y = 4

Answer 1

x2 +8x +4y +4 = 0

4y= -x2 -8x -4y = -.25*x^2 -2x -1

a = -.25b = -2c = -1

x position of vertex:

h = -b / 2a

h = 2 / 2*-.25h = 2 / -.5h = -4

y position of vertex:

k = ah^2 + bh + ck = -.25*-4^2 + -2*-4 + -1k = -4 +8 -1k = 3

VERTEX = (-4, 3)**************************************************************************

x value of focus =x value of vertex = -4

y value of focus =(1 (-b^2 -4ac)) / 4a

a = -.25 b = -2 c =-1

y value = (1 (-4 -4*-.25*-1)) / 4*-.25

y value = (1 (-4 -4*-.25*-1)) / -1

y value = (1 -4 +1) / -1y value = (-2 / -1)y value = 2

focus value = (-4, 2)

Answer is the last one.

Explanation:

3dgenuity

Answer 2

x^2+y^2-8x+4y+4=0 can be written as (x- 4)² + (y+2)²-16-4+4=0, or
(x- 4)² + (y+2)²=16, this is not a an equation of parabola, it is a circle's
its center is (4, -2) and its radius r=4