# Equation of a Parabola with Vertex Outside the Origin

Parabolas are conic sections that are obtained when a cone is cut by a plane. The plane that intersects the cone must be parallel to one lateral side of the cone. Parabolas are also defined as the set of points (x, y) that are equidistant from a line, called the directrix, and a fixed point, called the focus.

Here, we will learn to determine the equations of parabolas that have a vertex outside the origin. Also, we will learn how to graph parabolas when given the equations.

##### PRECALCULUS

Relevant for

Finding the equation of a parabola with vertex outside the origin.

See equations

##### PRECALCULUS

Relevant for

Finding the equation of a parabola with vertex outside the origin.

See equations

## Parabolas with vertex outside the origin

Parabolas do not always have their vertex at the point (0, 0). Many times, parabolas have their vertex outside the origin at the point $latex (h, k)$.

Recall that the equation of a parabola when the vertex is located at the origin is $latex {{x}^2}=4py$ or $latex {{y}^2}=4px$. Also, remember that the vertex form of a parabola is $latex y=a{{(x-h)}^2}-k$.

By combining these equations, we can obtain equations for parabolas with vertex outside the origin.

We start by solving for $latex {{(x-h)}^2}$:

$latex y=a{{(x-h)}^2}-k$

$latex {{(x-h)}^2}=\frac{1}{a}(y-k)$

Also, we have the equation $latex {{x}^2}=4py$. Comparing these two equations, we have $latex 4p=\frac{1}{a}$. Therefore:

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

If the parabola is oriented horizontally, the equation will be $latex {{(y-k)}^2}=4p(x-h)$. The h and k values always remain next to the x and y values respectively.

In the diagrams, we can see that, depending on which variable is squared, the parabola is oriented horizontally or vertically. When x is squared, the parabola is oriented vertically and when y is squared, the parabola is oriented horizontally.

Furthermore, if the value of p is greater than zero, the parabola opens upwards or to the right, that is, towards the positive part of the axes. On the other hand, when the value of p is negative, the parabola opens to the left or downwards, that is, towards the negative part of the axes.

## Graphs of parabolas with vertex outside the origin in standard form

Let’s analyze the equation $latex {{(y-3)}^2}=8(x+2)$. We have to start by finding the vertex, the axis of symmetry, the focus, and the directrix. Then, we will determine whether the parabola opens up, down, to the right, or to the left.

Because y is squared, we know that the parabola is oriented horizontally and opens to the right or to the left. Moreover, because p, in this case, 8 is positive, we know that the parabola must open to the right.

Using the general equation $latex {{(y-k)}^2} = 4p(x-h)$, we know that the vertex is the point $latex (h,~ k)$. In this case, the vertex is (-2, 3) and the axis of symmetry is $latex y = 1$.

Also, by forming the equation $latex 4p = 8$, we know that $latex p = 2$. Adding p to the x-value of the vertex, we have the focus, (0, 3). By subtracting p from the x-value of the vertex, we have the directrix, $latex x = -4$.

In addition to the critical points found, we have to determine a pair of symmetric points that are part of the curve to graph the parabola correctly. If we have $latex x = 6$, then we have $latex y = 11$ and $latex y = -5$. This means that both values ​​are part of the parabola.

## Examples with answers of parabolas with vertex outside the origin

The equations of parabolas that have their vertex outside the origin are used to solve the following exercises. Try to solve the exercises yourself before looking at the answer.

### EXAMPLE 1

What is the equation of the parabola that has a vertex at (-2, 4) and its directrix is $latex y = 7$?

Solution: We have to start by determining the orientation of the parabola. Because the directrix is horizontal, we know that the parabola will open up or down.

Also, we know that the directrix is located above the vertex, which means that the parabola opens downward and will be negative.

We use the vertex $latex (h, ~k)$ to find the value of . The equation for a horizontal directrix is $latex y = k-p$. Therefore, we have:

$latex 7=4-p$

$latex 3=-p$

$latex p=-3$

Using the general form, we have the following equation for the parabola:

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

$latex {{(x-(-2))}^2}=4(-3)(y-4)$

$latex {{(x+2)}^2}=-12(y-4)$

### EXAMPLE 2

Find the vertex, focus, axis of symmetry, and directrix of $latex {{(x + 3)}^2} = 4 (y + 4)$.

Solution: Comparing this equation with the general equation, we know that the vertex is $latex (-3, -4)$. Furthermore, we know that the parabola opens upwards since x is squared and p is positive.

We can form the equation $latex 4p=4$. By solving this, we have $latex p=1$. Therefore, the focus is $latex (-3, -4+1)=(-3, -3)$. The axis of symmetry is $latex x=-3$ and the directrix is $latex y=-3-1=-4$.

### EXAMPLE 3

Find the equation of the parabola that has a vertex at (-5, -1) and a focus at (-8, -1).

Solution: Since the vertex is $latex (-5, -1)$, we know that we have $latex h = -5$ and $latex k = -1$.

Also, the focus is (-8, -1), which means that the parabola will be horizontal since the coordinates in y of the vertex and the focus are the same. This means that p is subtracted or added from h.

The focus is equal to $latex (h+p, k)=(-8,-1)$. From this, we have the following:

$latex h+p=-8$

$latex -5+p=-8$

$latex p=-3$

Therefore, the equation of the parabola is:

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

$latex {{(y-(-1))}^2}=4(-3)(x-(-5))$

$latex {{(y+1)}^2}=-12(x+5)$