3.2.1: Geometric Description and Solutions of Two Particular Equations: the Circle and the Parabola

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Learning Objectives

By the end of this section, you will be able to:

Graph a circle with the origin at its center
Write the equation of a circle with the center at its origin
Graph a particular parabola
Write the equation of a particular parabola
Find the distance between two points
Find the midpoint of a line segment

Be Prepared

Before you get started, take this readiness quiz.

Check to see if (2,-3) is a solution to $x^2+y^2=7$.
Find the solution of $y=x^2$ of the form $(-2, \cdot)$.
Plot the point $(2,-3)$ on the coordinate plane.

Introduction to Quadratic Equations with Two Variables

A quadratic equation with two variables $x$ and $y$ is an equation that is equivalent to

\[Ax^2+By^2+Cx+Dy+Exy+F=0,\nonumber\]

where at least one of $A$ or $B$ is not zero.

In general, the graph of the solution to this type of equation is called a conic section (a circle, parabola, ellipse, hyperbola, line, two intersecting lines, or a point).

The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.

This figure shows two cones placed point to point. They are labeled nappes.

There are four conics—the circle, parabola, ellipse, and hyperbola and the degenerate ones also mentioned above. The next figure shows how the plane intersecting the double cone results in each curve.

Each of these four figures shows a double cone intersected by a plane. In the first figure, the plane is perpendicular to the axis of the cones and intersects the bottom cone to form a circle. In the second figure, the plane is at an angle to the axis and intersects the bottom cone in such a way that it intersects the base as well. Thus, the curve formed by the intersection is open at both ends. This is labeled parabola. In the third figure, the plane is at an angle to the axis and intersects the bottom cone in such a way that it does not intersect the base of the cone. Thus, the curve formed by the intersection is a closed loop, labeled ellipse. In the fourth figure, the plane is parallel to the axis, intersecting both cones. This is labeled hyperbola.

Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems.

We will discuss the solutions to this type of equation in certain cases. We'll treat with the case where $A=E=0$ or $B=E=0$ (i.e., where the equation is linear in one of the variables) and the case where $A=B$ and $E=0$ in which case, the graph is a circle.

The other cases could be treated in a similar way.

We start our discussion of conics with two particular important examples (and one family of examples).

A Particular Circle: Its geometric description and equation

We may be familiar with drawing a circle using a compass, or a pencil and a string. What is the figure that you draw this way: fix one end of a string on a paper. The place where this is fixed is called the center, and attach a pencil to the other end of the string. Then keeping the string taut, draw the curve that results from moving the pencil everywhere possible under the constraint of the string.

Let's suppose the length of the string is 1 unit. Then the distance between the center and any point on the drawn curve is 1 unit! Also, every point 1 unit away from the center must be part of the drawn curve. If we place our figure on a coordinate plane with the center at $(0,0)$ we have:

To find an equation which has this circle as its solution we label an arbitrary point on the circle $(x,y)$. We now form a right triangle as seen below.

Notice that in the picture the base of the triangle has length equal to the absolute value of the $x$-coordinate of the arbitrary point. Similarly, the height of the triangle is equal to the absolute value of the $y$-coordinate of the arbitrary point. Confirm by choosing this arbitrary point in different quadrants that this is true.

Now, Pythagoras' Theorem gives us that

\[ x^2+y^2=1\nonumber\]

which is the equation we are looking for! Every point on the circle satisfies this equation and every pair of numbers $(a,b)$ satisfying equation are the coordinates of a point distance 1 unit away from the center!

Notice that if the length of the string were $r$, then Pythagoras' Theorem would tell us

\[x^2+y^2=r^2.\nonumber\]

For such a circle, the center is $(0,0)$ and $r$ is called the radius.

So, the graph, following the geometric description, of the equation $x^2+y^2=1$ is

$desmos-graph (5).png$

More generally, the graph, following the geometric description, of the equation $x^2+y^2=r^2$ is

$desmos-graph (6).png$

Try It $\PageIndex{1}$

Graph the circle centered at $(0,0)$ with radius 1.

Answer: $desmos-graph (5).png$

Try It $\PageIndex{2}$

Graph the circle centered at $(0,0)$ with radius $7/2$.

Answer

Hint: Find four guiding points: above, below, left and right of the center (these are the points for which the grid is especially helpful!).

Check your answer on Desmos.

A Particular Parabola: A Geometric Description and Equation

Consider a line (called the directrix) and a point (called the focus) not on that line. The set of points that are the same distance to the focus as they are to the line (the length of the shortest line segment connecting the point to the line): see below.

The collection of points is called a parabola. Note that the point $(0,0)$ satisfies that condition: the distance from $(0,0)$ to the focus is $\dfrac{1}{4}$ as is the distance from $(0,0)$ to the directrix.

Let's consider a particular case and place it on a coordinate plane so that we can derive an equation for the collection of points satisfying the property above.

We will take (for later convenience) the focus to be $\left(0, \dfrac{1}{4}\right)$ and the directrix to be the horizontal line with equation $y=-\dfrac14$.

Then, as we did with the circle, we label an arbitrary point on the parabola with coordinates $(x,y)$. We see from the picture that the $y$ coordinate of the point is the distance to the $x$-axis and we have an additional distance 1 unit to the line! So the distance from the point to the line is $y+\dfrac14$. If the arbitrary point were in the second quadrant this would still be the case. It will be more convenient (but equivalent) to think of the parabola as the points whose distance squared is the same to the focus as it is to the directrix. The square of the distance to the line is then

\[\left(y+\dfrac14\right)^2=y^2+\dfrac12 y+\dfrac{1}{16}.\nonumber\]

To find the distance from $(x,y)$ to the focus $\left(0,\dfrac14\right)$ we look at the right triangle below and apply Pythagoras' Theorem.

We see the length of the base of the triangle is the absolute value of the $x$-coordinate of the point, i.e., $|x|$ (check to see this is true no matter where the point is), and the height of the triangle is $\left|y-\dfrac14\right|$. So, using Pythagoras' Theorem, we see that the square of the distance we are looking for is

\[x^2+\left|y-\dfrac14\right|^2\nonumber\]

and because $\left|y-\dfrac14\right|^2=\left(y-\dfrac14\right)^2$ the square of the distance from the point $(x,y)$ to the focus is

\[x^2+y^2-\dfrac12 y+\dfrac{1}{16}.\nonumber\]

Returning to the description of the parabola as the points which are the same distance to the focus as they are to the directrix, or, equivalently, so that the square of the distance from the point to the focus is the same as the square of the distance from the point to the directrix, we see

\[x^2+y^2-\dfrac12 y+\dfrac{1}{16}=y^2+\dfrac12 y+\dfrac{1}{16}\nonumber\]

or, by adding $-y^2+\dfrac12 y-\dfrac{1}{16}$ to both sides of the equation,

\[x^2=y\nonumber\]

or,

\[y=x^2.\nonumber\]

Every point on the parabola satisfies $y=x^2$ and every solution $(a,b)$ to this equation has the property that it is the same distance to the focus as it is to the directrix.

We can find some solutions to this particular equation by completing the table

$\begin{array}{cc}x & y \\ ---& --- \\-3 & \\ -2 & \\ -1 & \\ 0 & \\ 1 & \\ 2 & \\ 3 & \\ \end{array}$

by finding the values for $y$ that create solutions:

$\begin{array}{cc}x & y \\ ---& --- \\ -3 & 9 \\ -2 & 4 \\ -1 & 1 \\ 0 & 0 \\ 1 & 1\\ 2 & 4 \\ 3 & 9 \\ \end{array}$

This helps us get an idea of what the graph looks like:

$desmos-graph (4).png$

Try It $\PageIndex{3}$

Write the equation whose solutions are graphed here and verify by finding three solutions to your equation which are on the graph.

$desmos-graph (4).png$

Answer: The equation is $y=x^2$. The three examples will vary.

Try It $\PageIndex{4}$

Using the ideas presented in this subsection, graph the equation $x=y^2$.

Answer

Check your answer on Desmos.

$desmos-graph (7).png$

The Distance Formula

Here we mention, as we are discussing geometry in this section, the formula for the distance $d$ between to points and the midpoint of a line segment.

Consider two points on the coordinate plane $(x_1,y_1)$ and $(x_2,y_2)$. As we did above, we form the triangle shown below

The length of the base is $|x_2-x_1|$ and the height is $|y_2-y_1|$. We could do some examples with numbers to get a feel for these expressions. Putting these points in various quadrants will convince you of the need for the absolute values. Now, as before, we use Pythagoras' Theorem to obtain

\[|x_2-x_1|^2+|y_2-y_1|^2=d^2,\nonumber\]

or, equivalently,

\[(x_2-x_1)^2+(y_2-y_1)^2=d^2,\nonumber\]

so that the distance $d$ between $(x_1,y_1)$ and $(x_2,y_2)$ is

\[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.\nonumber\]

This is called the distance formula.

Similar Triangles and the Midpoint Formula

Two triangles are similar if they are the same shape (have the same angles). And if two triangles are similar, corresponding ratios of sides are equal.

The midpoint of the line segment connecting $(x_1,y_1)$ and $(x_2,y_2)$ is the point on the line segment which is the same distance from each of the endpoints.Look at the midpoint (label it $(a,b)$) of the line segment connecting $(x_1,y_1)$ and $(x_2,y_2)$ and form the triangle below. Here we don't draw the coordinate plane that this rests in so as to not clutter the picture.

$midpoint.jpg$

We see two triangles and since they are both right triangles and share an angle (since the sum of the measure of the interior angles of a triangle is 180 degrees, two angles are enough to determine the third angle), they are similar. Further, because $(a,b)$ is the midpoint, the length of the hypotenuse of the small triangle is half the length of the hypotenuse of the larger triangle. So, the same relation is true of the legs! It follows that $2(a-x_1)=x_2-x1$, or, solving for $a$, $a=\dfrac{x_1+x_2}{2}$. Similarly, $b=\dfrac{y_1+y_2}{2}$. By drawing the points in different relationships to each other, we can convince ourselves that these equations are true no matter where the points are located. So we have what is known as the midpoint formula. The midpoint of the line segment connecting $(x_1,y_1)$ and $(x_2,y_2)$ is

\[\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right).\nonumber\]

Example $\PageIndex{5}$

Find the midpoint of the line segment connecting $(-2,1)$ and $(3,5)$. Verify your answer by using the distance formula.

Solution

We find the midpoint by averaging the $x$ and $y$ components of the points (that is what the midpoint formula says):

The midpoint is $\left(\dfrac{-2+3}{2},\dfrac{1+5}{2}\right)=\left(\dfrac12,3\right).$

To verify that this is the midpoint, we will check that the distance between each end point and the proposed midpoint are the same and that this is half the length of the segment.

The distance between $(-2,1)$ and $\left(\dfrac12,3\right)$ is $\sqrt{\left(-2-\dfrac12\right)^2+(1-3)^2}=\sqrt{\dfrac{25}{4}+4}=\dfrac{\sqrt{41}}{2}$

and

The distance between $(3,5)$ and $\left(\dfrac12,3\right)$ is $\sqrt{\left(3-\dfrac12\right)^2+(5-3)^2}=\sqrt{\dfrac{25}{4}+4}=\dfrac{\sqrt{41}}{2}.$

The distance between $(-2,1)$ and $(3,5)$ is $\sqrt{\left(-2-3\right)^2+(1-5)^2}=\sqrt{25+16}=\sqrt{41}$

Because the midpoint is the point that is the same distance from the two endpoints and this distance is half the total distance, we have verified that $\left(\dfrac12,3\right)$ is the midpoint of the line segment connecting $(-2,1)$ and $(3,5)$.

Try It $\PageIndex{6}$

Find the midpoint of the line segment connecting $(2,-1)$ and $(-3,5)$. Verify your answer by using the distance formula.

Answer: The midpoint is $\left(-\dfrac12,2\right)$. The exposition of the verification will vary. The length of the line segment is $\sqrt{61}$.

Try It $\PageIndex{7}$

Find the midpoint of the line segment connecting $(3,-1)$ and $(-3,-5)$. Verify your answer by using the distance formula.

Answer: The midpoint is $\left(0,-3\right)$. The exposition of the verification will vary. The length of the line segment is $\sqrt{52}$.

A question for discussion is: Why isn't it enough to check any two of the three conditions we checked in the above example.

Applications

Example $\PageIndex{8}$

Find the center to a circle whose diameter has endpoints $(2,1)$ and $(-3,4)$.

Solution

The center of the circle is at the midpoint of this segment. So, to find the center we average the coordinates to find it is at

$(-1/2,5/2)$.

Try It $\PageIndex{9}$

Find the center to a circle whose diameter has endpoints $(-2,1)$ and $(3,5)$.

Answer

The center is at

$(1/2,3)$.

Example $\PageIndex{10}$

Use the distance formula to find the length of the hypotenuse of the triangle with vertices $(1,3)$, $(5,0$), and $(1,0)$.

Solution

We can see that the first point is above $(1,0)$ and the second point is to the right of $(1,0)$. The segments with $(1,0)$ as an enpoint form a right angle. Therefore the hypotenuse has endpoints $(1,3)$ and $(5,0$). So, to find the distance we look at the vertical component of the distance and the horizontal component of the distance: $3$ and $-4$ to see that

$d^2=3^2+(-4)^2=25$ so that $d=5$.

Alternatively, we can substitute $(1,3)$ for $(x_1,y_1)$ and $(5,0)$ for $(x_2,y_2)$ to see

\[d=\sqrt{(5-1)^2+(0-3)^2}=5.\nonumber\]

Try It $\PageIndex{11}$

Use the distance formula to find the length of the hypotenuse of the triangle with vertices $(4,-1)$, $(2,1$), and $(2,-1)$.

Answer: The hypotenuse has length $2\sqrt{2}$.

Writing Exercises $\PageIndex{12}$

Explain reasoning behind the midpoint formula.
Can you describe what a circle is using the words: center, radius and distance?
How is the distance formula related to the Pythagorean Theorem?
How can you recognize solutions to a quadratic equation in two variables as forming a line, a parabola, or a circle?
What is the value of finding 4 points on the circle when graphing by hand?
What is a perpendicular bisector?
What information about a line is convenient to deal with if we want to write it’s equation in slope-intercept form?
What about point-slope form?

Exit Problem $\PageIndex{13}$

Identify the center and the radius of the circle given by the equation $x^2 -4x +y^2+6y-23 =0.$

Graph the circle and label four points on it.

Key Concepts

The circle with center $(0,0)$ and radius $r$ has the equation $x^2+y^2=r^2$
A particular parabola has the equation $y=x^2$ and its graph is \[ \] $desmos-graph (4).png$ .
The distance $d$ between $(x_1,y_1)$ and $(x_2,y_2)$ is
\[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.\]
The midpoint of the line segment connecting $(x_1,y_1)$ and $(x_2,y_2)=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right),$ or in words the coordinates of the midpoint is the average of the coordinates.

Search

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