Determine the equation of the porabola with vertex (1,8) and focus (1,15)

Answer 1

Notice first that the points (1,8) and (1,15) form a line parallel to the y-axis, and our focus is higher than our vertex, so our parabola opens upwards. Suppose the equation of the parabola is

y = a(x-k)²+h.

The vertex is defined as the minimum or maximum point which is (k,h) (think of the parabola y=x² whose vertex is (0,0)). So k = 1, h = 8. Now the focus is defined as the point

F = (k,h+1/4a)

But we know F, which is (1,15), so 15 = 8+1/4a which means 4a=1/7 or a = 1/28. So our final parabola is

y = 1/28 * (x-1)²+ 8.

Answer 2

V(1,8) F(1,15)

with focus above the vertex, it's an upward opening parabola, with a>0

axis of symmetry: x=1

y= a(x-h)^2 + k

y =a(x-1)^2 +8

y= (1/28)(x-1)^2 + 8

focus is (1,15) directrix is y=1, vertex is half way between focus and directrix.

any point on the parabola is also equidistant from both. such as the point (15,15)

a distance of 14 from each

15 = (1/28)(14)^2 + 8 = 7+8=15

Answer 3

These are the standard conic section equations of a parabola with vertex (h, k) and distance from vertex to focus, p:

(x - h)² = 4p(y - k) for a vertical parabola or

(y - k)² = 4p(y - k) for a horizontal parabola

*I have seen books use alternative forms where the 4p is on the other side of the equation like so:

1/(4p)(x - h)² = (y - k) but to avoid the fraction I'll stick with the above forms.

Since we have a vertex at (1, 8) and focus at (1, 15) we can picture our parabola opening towards the focus which is above the vertex so we have a parabola opening upwards. Thus, we'll use the top form. Since (1, 15) is 7 units above the vertex at (1, 8) we know p = 7. Substituting our vertex and p into the form we end up with:

(x - 1)² = 28(y - 8)

That should be your final answer.