How do you find the standard equation given focus (8,10), and vertex (8,6)?

Question

Kennedy Adams · Accepted Answer

The focus is above the vertex, therefore, the vertex form of the equation is:

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

Use the focus to compute #a = 1/(4f)#

Expand the equation into standard form.

The distance from the vertex to the focus is known as the focal distance, or f: #f = 10-6# #f = 4# Determine "a"'s value: #a = 1/(4f)# #a = 1/16# The vertex tells us that #h = 8 and k = 6# Putting these values in the vertex form instead of: #y = 1/16(x-8)^2+6# Enlarge the square: #y = 1/16(x^2-16x+64)+6# #y = x^2/16-x+4+6# #y = x^2/16-x+10# This is standard form.

Ethan Adkins · Answer

Find the distance between vertex and focus. Call that p.
Since it opens upward, p >0. Use (x-h)2 = 4p(y - k).

The equation of the parabola that opens up or down and has vertex (h, k) is #(x-h)^2 = 4p(y - k)# where p is the difference between the y-coordinates of the focus and the vertex. Since (h, k) = (8, 6) and p = 10 - 6 = 4 in this example, #(x-8)^2 = 16(y - 6)# [If we want this in the "standard form," that usually means solving for the variable that is not squared. Distribute the 16, and add...] This might have been the form you were looking for. #(x-8)^2 = 16(y - 6)# #(x-8)^2 = 16y - 96# #(x-8)^2 + 96 = 16y# #y = 1/16 (x-8)^2 + 6# ] If you were only interested in the standard form, set a = 1/(4p) and go straight to the vertex form: Since a = 1/(4*4) = 1/16, #y = a(x-h)^2 + k# That is, #y = 1/16 (x-8)^2 + 6# Use FOIL and distribute if you prefer the form #y = ax^2 + bx + c# I won't ruin the enjoyment for you.

Ethan Adkins · Answer

undefined To find the standard equation of a parabola given the focus and vertex, we first need to determine if the parabola opens upwards or downwards.

If the parabola opens upwards, then the vertex is the lowest point on the graph, and the focus is above the vertex.

If the parabola opens downwards, then the vertex is the highest point on the graph, and the focus is below the vertex.

Since the focus is given as (8, 10) and the vertex as (8, 6), we can see that the focus is above the vertex. Therefore, the parabola opens upwards.

The standard equation of a parabola that opens upwards and has its vertex at (h, k) is given by:

$$ (x - h)^2 = 4p(y - k) $$

where (h, k) is the vertex and p is the distance from the vertex to the focus (and also the distance from the vertex to the directrix).

Since the focus is (8, 10) and the vertex is (8, 6), the distance from the vertex to the focus is 4 units (10 - 6 = 4).

So, the standard equation of the parabola is:

$$ (x - 8)^2 = 4(4)(y - 6) $$

$$ (x - 8)^2 = 16(y - 6) $$