How do you write the equation of the parabola in vertex form given Vertex (2,4), Focus (2,6)?

Question

How do you write the equation of the  parabola in vertex form given Vertex (2,4), Focus (2,6)?

Eli Assouline · Accepted Answer

#(x-2)^2 = 8(y-4)#

Standard equation of a vertical parabola with vertex (h,k) is #(x-h)^2 =4p(y-k)# where p is the distance of the focus from the vertex. In the present case vertex is (2,4). Since the focus is at (2,6), the distance of the focus from vertex , that is 'p' would be 6-4= 2 units.

The required equation of the parabola thus can be written as :

#(x-2)^2 = 8(y-4)#

This is shown in the figure below:

The required equation of the parabola thus can be written as :

#(x-2)^2 = 8(y-4)#

This is shown in the figure below:

Eli Assouline · Answer

undefined The equation of a parabola in vertex form, given the vertex $(h, k)$ and the focus $(h, k + p)$, where $p$ is the distance between the vertex and the focus along the axis of symmetry, can be written as:

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

In this case, the vertex is $(2, 4)$ and the focus is also at $(2, 6)$. Since the vertex and focus have the same x-coordinate, the parabola opens either upwards or downwards.

The distance between the vertex and the focus is $ p = 6 - 4 = 2 $.

Now substitute the values into the vertex form equation:

$$ (x - 2)^2 = 4 \cdot 2(y - 4) $$

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

This is the equation of the parabola in vertex form.