Find the standard form of the equation of the parabola with the given characteristics. Vertex: (4, -1); focus: (4, -4)

Answer 1

the focus is below the vertex, so the parabola is downward opening with

standard equation: y= a(x-4)^2 -1

where a<0

to calculate a, plug in another point on the parabola

(10,-4) which is equidistant from the focus and directrix y=2, 6 from each. it may help to skecth the parabola, directrix, focus and vertex

-4 = a(10-4)^2 -1

-4 =36a -1

36a = -3

a = -3/36=-1/13

y = (-1/13)(x-4)^2 -1 is the parabola's equation in vertex form

y =(-1/13)x + (8/13)x -29/13 is the parabola's equation in standard form,

y=ax^2+bx+c, a=-1/13, b=8/13, c =-29/13