# Parabolas as Conic Sections

### Parts of a Parabola

A parabola is a conic section defined in terms of distance to a fixed point and a fixed line.

The graphs of quadratic functions are parabolas. However, a parabola is defined differently when it is considered as a conic section. A parabola as a conic section is the set of points such that the distance from a fixed point, the focus, is equal to the distance to a fixed line, the directrix.

The properties of a parabola include:

• For any point on a parabola, the distance from the point to the focus is equal to the distance from that same point to the directrix.
• The vertex of a parabola is the midpoint of the segment from the focus to the directrix.
• The axis of symmetry of the parabola divides the parabola into two halves that are mirror images. It passes through both the focus and the vertex.
A parabola can have its vertex at the origin or elsewhere on the coordinate plane, and its axis of symmetry may be vertical or horizontal. The equation of a parabola can be written in terms of the focus and directrix. A parabola with a vertical axis of symmetry and its vertex at the origin has an equation that can be written in the form:
$y=\left(\frac{1}{4a}\right)x^2$
It can be rewritten as:
$x^2=4ay$
• If $a>0$, the parabola opens upward.
• If $a<0$, the parabola opens downward.
• The focus is at $(0,a)$, and the directrix is at $y=-a$.

### Parabolas with a Vertical Axis of Symmetry

Opens Upward Opens Downward
$x^2=8y$ $x^2=-8y$
Focus: $(0,2)$ Focus: $(0,-2)$
Directrix: $y = -2$ Directrix: $y=2$

For the parabola $x^2=8y$ or $x^2=4(2)y$, the value of $a$ is 2. Since $a$ is positive, the parabola opens upward and its focus is above the vertex. For the parabola $x^2=-8y$ or $x^2=4(-2)y$, the value of $a$ is –2. Since $a$ is negative, the parabola opens downward, and its focus is below the vertex. Both parabolas have a vertical axis of symmetry and a horizontal directrix.

A parabola with a horizontal axis of symmetry and its vertex at the origin has an equation that can be written in the form:
$x=\left(\frac{1}{4a}\right)y^2$
It can also be rewritten as:
$y^2=4ax$
• If $a>0$, the parabola opens to the right.
• If $a<0$, the parabola opens to the left.
• The focus is at $(a,0)$, and the directrix is at $x=-a$.

### Parabolas with a Horizontal Axis of Symmetry

Opens Right Opens Left
$y^2=8x$ $y^2=-8x$
Focus: $(2,0)$ Focus: $(-2,0)$
Directrix: $x=-2$ Directrix: $x=2$

For the parabola $y^2=8x$ or $y^2=4(2)x$, the value of $a$ is 2. Since $a$ is positive, the parabola opens to the right, and its focus is to the right of the vertex. For the parabola $y^2=-8x$ or $y^2=4(-2)x$, the value of $a$ is –2. Since $a$ is negative, the parabola opens to the left, and its focus is to the left of the vertex. Both parabolas have a horizontal axis of symmetry and a vertical directrix.

### Graphing Parabolas

A parabola with a given equation can be graphed in the coordinate plane by locating the vertex, focus, and directrix.
A parabola may or may not have its vertex at the origin. A parabola with a vertex at the point (h, k) represents a horizontal translation by h units and a vertical translation by k units of a parabola with its vertex at the origin.

### Equations of Parabolas with Vertex at $(h,k)$

Equation Axis of Symmetry Focus Directrix
$(x-h)^2=4a(y-k)$
Vertical: $x=h$
$(h,k+a)$
$y = k-a$
$(y-k)^2=4a(x-h)$
Horizontal: $y=k$
$(h+a,k)$
$x = h-a$

Step-By-Step Example
Graphing a Parabola
A parabola has the equation:
$8(y-2)=(x+5)^2$
Calculate the vertex, focus, and directrix of the parabola. Sketch a graph of the parabola.
Step 1

Write the equation in standard form.

The squared variable in the equation is $x$, so the parabola has a vertical axis of symmetry and the standard form of its equation is:
$(x-h)^2=4a(y-k)$
Rewrite the equation using the standard form:
\begin{aligned}8(y-2)&=(x+5)^2\\(x+5)^2&=8(y-2)\\(x-(-5))^2&=8(y-2)\\(x-(-5))^2&=4(2)(y-2)\end{aligned}
Step 2
Use the standard form of the equation to identify the values of $h$, $k$, and $a$.
\begin{aligned}(x-h)^2&=4a(y-k)\\(x-(-5))^2&=4(2)(y-2)\end{aligned}
So, $h=-5$, $k=2$, and $a=2$.
Step 3

Determine the vertex, focus, and directrix.

The vertex $(h, k)$ is $(-5, 2)$.

The coordinates of the focus are $(h, k+a)$.

The focus is $(-5, 2+2)$, or $(-5, 4)$.

The equation of the directrix is:
$y = k-a$
The directrix is the horizontal line:
$y=2-2=0$
Step 4

Plot the focus and vertex, and graph the directrix.

To locate another point on the parabola, draw a square with one vertex at the focus and one side along the directrix. The vertex $(-1, 4)$ of this square is a point on the parabola. Use symmetry to locate the point $(-9,4)$.
Solution
Draw the parabola through the three points.