Question

What is an equation of a parabola with the given vertex and focus?

vertex: (-2,5); focus: (-2,6)

I really need someone to explain how to do this. I am so confused.

Answer 1

Since the vertex and focus both lie on the line x = =2, the axis of symmetry is horizontal, meaning the equation is

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

Since the focus above the vertex, p is positive.

You know that the parabola x^2 = 4py has the vertex at a distance p from the focus. Here, that distance is 1, so p = 1/4. So, your equation is

(x+2)^2 = 4(1)(y-5)
(x+2)^2 = 4(y-5)

See

http://www.wolframalpha.com/input/?i=parabola+(x%2B2)%5E2+%3D+4(y-5)

Answer 2

As you noticed, the axis is vertical, on he line x = -2

Answer 3

I see other typos. The final answer is correct, but let me just do it right:

Since the vertex and focus both lie on the line x = -2, the axis of symmetry is vertical, meaning the equation is

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

Since the focus above the vertex, p is positive.

You know that the parabola x^2 = 4py has the vertex at a distance p from the focus. Here, that distance is 1, so p = 1. So, your equation is

(x+2)^2 = 4(1)(y-5)
(x+2)^2 = 4(y-5)

Answer 4

where does the 5 in "(y-5)" come from 💀