# Conics Part 1: Circles and Parabolas

Note that Part 2 of Conics (Ellipses, Hyperbolas, and Identifying the Conic) is here.

Conics (circles, ellipses, parabolas, and hyperbolas) involves a set of curves that are formed by intersecting a plane and a double-napped right cone (probably too much information!). But in case you are interested, there are four curves that can be formed, and all are used in applications of math and science:

In the Conics section, we will talk about each type of curve, how to recognize and graph them, and then go over some common applications. Always draw pictures first when working with Conics problems!

## Table of Conics

Before we go into depth with each conic, here are the Conic Section Equations. Note that you may want to go through the rest of this section before coming back to this table, since it may be a little overwhelming at this point!

Note: The standard form (general equation) for any conic section is:

$\begin{array}{l}A{{x}^{2}}+Bxy+C{{y}^{2}}+Dx+Ey+F=0,\,\,\,\,\\\,\,\,\,\,\,\,\,\,\,\,\text{where}\,\,\,A,B,C,D,E,F\text{ are constants}\end{array}$

It actually turns out that, if a conic exists, if ${{B}^{2}}-4AC<0$, it is a circle or ellipse, if ${{B}^{2}}-4AC=0$, it is a parabola, and if ${{B}^{2}}-4AC>0$, it is a hyperbola.

Note: We can also write equations for circles, ellipses, and hyperbolas in terms of cos and sin, and other trigonometric functions using Parametric Equations; there are examples of these in the Introduction to Parametric Equations section.

## Circles

You’ve probably studied Circles in Geometry class, or even earlier. Circles are defined as a set of points that are equidistant (the same distance) from a certain point; this distance is called the radius of a circle.

Here is the equation for a circle, where $r$ is the radius: $\displaystyle \begin{array}{l}\text{Center }\left( {0,\,\,0} \right):\,\,\,\,{{x}^{2}}+{{y}^{2}}={{r}^{2}}\\\text{Center }\left( {h,\,\,k} \right):\,\,\,\,{{\left( {x-h} \right)}^{2}}+{{\left( {y-k} \right)}^{2}}={{r}^{2}}\end{array}$.

If we were to solve for $y$ in terms of $x$ (for example, to put in the graphing calculator), we’d get:  $\displaystyle \begin{array}{l}\text{Center }\left( {0,\,\,0} \right):\,\,\,\,y=\pm \sqrt{{{{r}^{2}}-{{x}^{2}}}}\\\text{Center }\left( {h,\,\,k} \right):\,\,\,\,y=\pm \sqrt{{{{r}^{2}}-{{{\left( {x-h} \right)}}^{2}}}}+k\end{array}$  .

Here are graphs of sample circles, with their domains and ranges:

 Circle with Center $\boldsymbol{\left( {0,0} \right)}$ Circle with Center $\boldsymbol{\left( {h,k} \right)}$ Domain: $\left[ {-2,2} \right]$      Range: $\left[ {-2,2} \right]$ Domain: $\left[ {-3,7} \right]$      Range: $\left[ {-8,2} \right]$

Sometimes we have to complete the square to get the equation for a circle. We learned how to complete the square with quadratics here in the Factoring and Completing the Square section. Note that, after completing the square, we may not necessarily get a circle if the coefficients of ${{x}^{2}}$ and ${{y}^{2}}$ are not both positive 1, as we’ll see later.

Here are some examples:

### Writing Equations of Circles

Sometimes you will have to come up with the equations of circle, or equations of tangents of circles.

Problem:

Write the equation of the line that is tangent to the circle ${{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=61$ at the point $\left( -2,-8 \right)$.

Solution:

A line tangent to a circle means that it touches the circle at one point on the outside of the circle, at a radius that is perpendicular to that line:

For this problem, since we only have one point on the tangent line $\left( {-2,-8} \right)$, we’ll have to get the slope of the line to get its equation. Remember here in the Coordinate System and Graphing Lines that perpendicular lines have slopes that are opposite reciprocals of each other.

Let’s draw a picture, and then get the solution:

Here’s another type of problem you might see:

Problem:

The lines $\displaystyle y=\frac{4}{3}x-\frac{5}{3}$ and $\displaystyle y=-\frac{4}{3}x-\frac{{13}}{3}$ each contain diameters of a circle, and the point $\left( {-5,0} \right)$ is also on that circle. Find the equation of this circle.

Solution:

If two lines are both diameters of the same circle, where they intersect must be the center of the circle. In this case, it was easier to draw a picture to see that this is true:

Now we can get the center of the circle by finding the intersection of the two lines. Since we have another point, too, we can get the equation of the circle:

## Applications of Circles

Problem:

A pizza delivery area can be represented by a circle, and extends to the points $\left( {0,18} \right)$ and $\left( {-6,8} \right)$ (these points are on the diameter of this circle). Write an equation for the circle that models this delivery area.

Solution:

If we draw a picture, we’ll see that we’ll have to use both the Distance Formula and Midpoint Formula from the Coordinate System and Graphing Lines section:

## Parabolas

Let’s revisit parabolas (a type of quadratic), but go into a little more depth here. We studied Parabolas in the Introduction to Quadratics section, but we only looked at “vertical” parabolas with positive coefficients (“cup up”) and negative coefficients (“cup down”).

Vertical parabolas are in the form $y=a{{\left( x-h \right)}^{2}}+k$, where $\left( {h,\,k} \right)$ is the vertex and $x=h$ is the axis of symmetry or line of symmetry (LOS). Note that this can also be written $y-k=a{{\left( {x-h} \right)}^{2}}$ or $b\left( {y-k} \right)={{\left( {x-h} \right)}^{2}}+k$, where $\displaystyle b=\frac{1}{a}$. The line of symmetry (LOS) is a line that divides the parabola into two parts that are mirror images of each other.

Parabolas can also be in the form $x=a{{\left( y-k \right)}^{2}}+h$, where $\left( {h,\,k} \right)$ is the vertex, and $y=k$ is the LOS; this is a horizontal parabola. In these cases, parabolas with a negative coefficient faces left.

Technically, a parabola is the set of points that are equidistant from a line (called the directrix) and another point not on that line (called the focus, or focal point). For vertical (up and down) parabolas, the directrix is a horizontal line (“$y=$”), and for horizontal (sideways) parabolas, the directrix is a vertical line (“$x=$”).

If $p$ is the distance from the vertex to the focus point (called the focal length), it is also the distance from the vertex to the directrix. This makes the distance from the focus to the directrix is $2p$. Note that the focus is always “inside” the parabola on the line of symmetry, and the directrix is “outside” the parabola.

Also note that the line perpendicular to the line of symmetry (and thus parallel to the directrix) that connects the focus to the sides of the parabola is called the latus chord, latus rectum, focal chord or focal rectum; the length of this chord (focal width or focal diameter) is $4p$. This is cool! To draw the parabola, if you know $p$, you can just go out $2p$ on either side of the focus to get more points!

Here is a vertical parabola with center $\left( {0,\,0} \right)$:

If the vertex is at the origin $\left( {0,\,0} \right)$, the equation of a vertical parabola is  $y=a{{x}^{2}}$, and $\displaystyle a=\frac{1}{{4p}}$; if you do the algebra, it follows that that $\displaystyle p=\frac{1}{4a}$. For example, if $p=4$ (length of focus to vertex), the equation of the parabola would be $\displaystyle y=\frac{1}{{4\left( 4 \right)}}{{x}^{2}}\,=\frac{1}{{16}}{{x}^{2}}$.

Here are the four different “directions” of parabolas and the generalized equations for each. It looks complicated, but it’s really not that bad; just remember to draw the parabolas, and you’ll get the hang of it pretty quickly. Also remember that the “$h$” always goes with the “$x$” and the “$k$” always goes with the “$y$”.

Note that in these formulas and subsequent examples, I assume $p$ is a distance; thus, it is always positive (and so is $\displaystyle a=\frac{1}{{4p}}$). Many teach using just one formula for positive and negative coefficients; thus, $p$ is just the coefficient, and can be negative. Sorry for the confusion; either approach can be used.

Note that sometimes (as in the problem below) we have to complete the square to get the equation in parabolic form; we did this here in the Solving Quadratics by Factoring and Completing the Square Section. Let’s do some problems!

### Writing Equations of Parabolas

Here are some problems where we need to find equations of parabolas, given certain conditions.

Here are more problems where we need to find an equation for a parabola; the second one is really tricky!

## Applications of Parabolas

The main application of parabolas, like ellipses and hyperbolas, are their reflective properties (lines parallel to the axis of symmetry reflect to the focus). They are very useful in real-world applications like telescopes, headlights, flashlights, and so on.

Problem:

The equation  $\displaystyle \frac{1}{32}{{x}^{2}}$ models cross sections of parabolic mirrors that are used for solar energy. There is a heating tube located at the focus of each parabola; how high is this tube located above the vertex of its parabola?

Solution:

For problems like these, unless otherwise noted, just assume the vertex of the parabola is at $\left( {0,0} \right)$. Since we know that the equation of a parabola is $y=a{{x}^{2}}$, where  $\displaystyle a=\frac{1}{4p}$ and $\displaystyle p=\frac{1}{4a}$, then for $\displaystyle \frac{1}{32}{{x}^{2}}$, we have $\displaystyle \frac{1}{32}=\frac{1}{4p}$. We can either cross multiply or just do mental math to see that $p=8$. The heating tube needs to be located 8 units above the vertex of the parabola.

Problem:

A searchlight has a parabolic reflector (has a cross section that forms a “bowl”). The parabolic “bowl” is 16 inches wide from rim to rim and 12 inches deep. The filament of the light bulb is located at the focus. (a) What is the equation of the parabola used for the reflector?  (b) How far from the vertex is the filament of the light bulb?

Solution:

Let’s graph this particular parabola, again putting the vertex at $\left( {0,0} \right)$.

Problem:
The bulb in a searchlight is placed at the focus of a parabolic mirror, which is .75 feet from the vertex. This causes the rays of light from the bulb to bounce off of the mirror as parallel rays, providing a concentrated beam of light.

(a) What is the equation of the parabola, if the focal diameter of the bulb is 3 feet?  (b) Suppose, in another searchlight, the focal diameter is 4 feet. If the depth of both searchlights is 5 feet, how much greater is the width of the opening of this larger light than the searchlight with a focal diameter of 3 feet?

Solution:
Let’s graph this particular parabola, again putting the vertex at $\left( {0,0} \right)$. As in other problems, it’s easier to make a vertical parabola.

Problem:

The cables of the middle part of a suspension bridge are in the form of a parabola, and the towers supporting the cable are 600 feet apart and 100 feet high. What is the height of the cable at a point 150 feet from the center of the bridge?

Solution:

Let’s draw a picture of the bridge, and place the middle of the cable (vertex) at the point $\left( {0,0} \right)$.

For Practice: Use the Mathway widget below to try a problem. Click on Submit (the blue arrow to the right of the problem) and click on Find the Circle Using the Diameter End Points, for example, to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

Scroll to Top