Since the vertex is (-2,5), the equation must be of the form:
y = a(x + 2)^2 + 5
where a is a constant that determines the shape of the parabola. To find a, we need to use the fact that the focus is at (-2,6). Recall that the focus lies on the axis of symmetry of the parabola, which is a vertical line passing through the vertex. In this case, the equation of the axis of symmetry is x = -2.
The distance from the vertex to the focus is given by the formula:
p = 1/4a
where p is the distance from the vertex to the focus. We know that p = 1, since the focus is above the vertex by a distance of one unit. Substituting this into the formula and solving for a, we get:
1 = 1/4a
a = 4
Therefore, the equation of the parabola is:
y = 4(x + 2)^2 + 5
What is an equation of a parabola with the given vertex and focus?
vertex: (-2, 5) focus: (-2,6)
2 answers
AAAaannndd the bot gets it wrong yet again!
Your parabola has focus at (-2,81/16)
recall that the parabola
x^2 = 4py
has vertex at (0,0) and focus at (0,p)
So our parabola has p=1, making its equation
(x+2)^2 = 4(y-5)
or
y = 1/4 (x+2)^2 + 5
Your parabola has focus at (-2,81/16)
recall that the parabola
x^2 = 4py
has vertex at (0,0) and focus at (0,p)
So our parabola has p=1, making its equation
(x+2)^2 = 4(y-5)
or
y = 1/4 (x+2)^2 + 5