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# 6.2: Circles and Ellipses

Created by: CK-12

## Learning Objectives

• Understand the difference between an “oval” and an ellipse.
• Recognize and work with equations for ellipses.
• Derive the focal property of ellipses.
• Understand the equivalence of different definitions of ellipses.
• Reconstruct Dandelin’s sphere construction.
• Know some of the different ways people have approached ellipses throughout history.
• Understand some of the important applications of ellipses.

## Introduction

Let’s begin with the first class of shapes discussed in the last section. When the plane makes a finite intersection with one side of the cone, we get either a circle or the “oval-shaped” object illustrated in the previous section. It turns out that this is no ordinary oval, but something called an ellipse, a shape with special properties.

Like parallelograms, or any other shape with lots of interesting properties, ellipses can be defined by some of these properties, and then the other properties necessarily follow from the definition. For example, a parallelogram is typically defined as a quadrilateral with each pair of opposite pairs of sides parallel. Once you define it this way, it follows that the opposite sides must also be equal in length, and that the diagonals must bisect each other. Well, if you instead started by defining a parallelogram by one of these other properties, for instance opposite sides having equal lengths, then you would end up with the same class of shapes. The same thing happens for ellipses. One way to define an ellipse is as a “stretched out circle”. It’s the shape you would get if you sketched a circle on a deflated balloon and then stretched out the balloon evenly in two opposite directions:

It’s also the shape of the surface of water that results when you tilt a round glass:

Or an ellipse could be thought of as the shape of a circle drawn on a piece of paper when it is viewed at an angle.

## Equations of Ellipses

This “stretching” can be represented algebraically. For simplicity, take the circle of radius 1 centered at the origin (0,0). The distance formula tells us that this is the set of points $(x, y)$ that is a distance 1 unit away from the origin.

$D & = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\1 & = \sqrt{(x - 0)^2 + (y - 0)^2} \\ 1 & = x^2 + y^2$ This equation, $x^2 + y^2 = 1$, can be altered to stretch the circle in the horizontal (i.e. $x-$axis) direction by dividing the $x$ variable by a constant $a>1$,

$\left (\frac{x}{a} \right )^2 + y^2 = 1$

Why does this stretch the circle horizontally? Well, the effect of dividing $x$ by $a$ is that for each $y-$value in an ordered pair $(x, y)$ that satisfies the original equation, the corresponding $x$ value must be multiplied by $a$ in order for the pair to make a solution to the altered equation. So solutions $(x, y)$ of the circle are in one-to-one correspondence with solutions $(ax, y)$ of the altered equation, hence stretching the corresponding graph to the left and right by a factor of $a$. For example, here is the graph of $\left (\frac{x}{2} \right )^2 + y^2 = 1$:

Generalizing the equation by allowing a stretch in the vertical direction, we get the following.

$\left (\frac{x}{a} \right )^2 + \left (\frac{y}{b} \right )^2 =1$

The factor $a$ stretches the circle in the horizontal direction and the factor $b$ stretches the circle in the vertical direction. If $a=b$, this is just a circle. When $a \neq b$, this equation represents an ellipse. The ellipse is stretched in the horizontal direction if $b < a$ and it is stretched in the vertical direction if $a < b$. Often the above equation is written as follows.

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

This is called the standard form of the equation of an ellipse, assuming that the ellipse is centered at (0,0).

To sketch a graph of an ellipse with the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, start by plotting the four axes intercepts, which are easy to find by plugging in 0 for $x$ and then for $y$. Then sketch the ellipse freehand, or with a graphing program or calculator.

Example 1 Sketch the graph of $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

Solution: This equation can be rewritten as $\frac{x^2}{2^2} + \frac{y^2}{3^2} = 1$. After setting $x = 0$ and $y = 0$ to find the $y-$ and $x-$intercepts, (0,3), (0,-3), (2,0), and (-2,0), we sketch the ellipse about these points:

Example 2 Sketch the graph of $\frac{x^2}{16} + y^2 = 1$.

Solution: This can be rewritten as $\frac{x^2}{4^2} + \frac{y^2}{1^2} = 1$. After finding the intercepts and sketching the graph, we have:

The segment spanning the long direction of the ellipse is called the major axis, and the segment spanning the short direction of the ellipse is called the minor axis. So in the last example the major axis is the segment from (-4,0) to (4,0) and the minor axis is the segment from (0,-1) to (0,1).

The major and minor axes are examples of what are sometimes called reference lines. Apollonius, the Ancient Greek mathematician who wrote an early treatise on conics, used these and other reference lines to orient conic sections. Though the Greeks did not use a coordinate plane to discuss geometry, these reference lines offer a framing perspective that is similar to the Cartesian plane that we use today. Apollonius’ way of framing conics with reference lines was the closest mathematics came to the system of coordinate geometry that you know so well until Descartes’ and Fermat’s systematic work in the seventeenth century.

Example 3 Not all equations for ellipses start off in the standard form above. For example, $25x^2 + 9y^2 = 225$ is an ellipse. Put it in the proper form and graph it.

Solution: First, divide both sides by 225, to get: $\frac{x^2}{9} + \frac{y^2}{25} = 1$, or $\frac{x^2}{3^2} + \frac{y^2}{5^2} = 1$. Finding the intercepts and graphing, we have:

### Review Questions

1. It was mentioned above that when a round glass of water is tilted, the surface of the water is an ellipse. Using our working definition of an ellipse as “stretched out circle”, explain why you think the water takes this shape.
2. Sketch the following ellipse: $36x^2 + 25y^2 = 900$
3. Now try sketching this ellipse where the numbers don’t turn out to be so neat: $3x^2 + 4y^2 = 12$

1. Answers may vary, but should explain why the shape that results stretches a circle in one direction because the width of the glass is constant.

## The Focal Property

In every ellipse there are two special points called the foci (foci is plural, focus is singular), which lie inside the ellipse and which can be used to define the shape. For an ellipse centered at (0,0) that is wider than it is tall, its major axis is horizontal and its foci are at $\left ( \sqrt{a^2 - b^2},0 \right )$ and $\left ( - \sqrt{a^2 - b^2},0 \right )$.

What is the significance of these points? The ellipse has a geometric property relating to these points that is similar to a circle’s relationship with its center. Remember a circle can be thought of as the set of points in a plane that are a certain distance from the center point. In fact, that is typically the definition of a circle. Well, the foci act like the center except that there are two of them. An ellipse is the set of points where the sum of the distance between each point on the ellipse and each of the two foci is a constant number. In the diagram below, for any point $P$ on the ellipse, $F_1 P + F_2P = d$, where $F_1$ and $F_2$ are the foci and $d$ is a constant.

## Technology

This definition using the sum of the focal distance gives us a great way to draw ellipses. Sure, you can just graph them on your calculator. But why not utilize a simpler technology that does the job just as well? As you know, a circle could be drawn by fixing a string to a piece of paper, tying the other end to a pencil, and then drawing the curve that keeps the string taut.

Similarly, an ellipse can be drawn by taking a string that is longer than the distance between two points, fixing the two ends of the string to the two points. Then drawing all points that can be drawn when the string is taut and the pencil is touching it. In the diagram above, the dotted line represents the string of fixed length, which is attached at the foci $F_1$ and $F_2$. The string is looped around the pencil at $P$, and then the pencil is moved, keeping the string taut, drawing all the points whose sum of distances between $F_1$ and $F_2$ is the length of the string.

### Review Questions

1. Use string and tacks to draw ellipses that
1. are nearly circles
2. are very different than circles

For each of these, what can you say about how the distance between the foci and the length of the string compare?

1. Drawings may vary. For ellipses that are nearly circles, the distance between the foci is small compared to the length of string.

The foci can also be used to measure how far an ellipse is “stretched” from a circle. The symbol $\varepsilon$ stands for the eccentricity of an ellipse, and it is defined by the distance between the foci divided by the length of the major axis, or $\frac{\sqrt{a^2 - b^2}}{a}$ for horizontally oriented ellipses and $\frac{\sqrt{b^2 - a^2}}{b}$ for vertically oriented ellipses. Since a circle is an ellipse where $a = b$, circles have an eccentricity of 0.

### Review Questions

1. What is the full range of the eccentricity of an ellipse? What does it look like near the extremes of this range?

1. The interval of possible values is $\varepsilon \in [0,1)$. At $\varepsilon = 0$, the ellipse is a circle; as the eccentricity approaches 1 it becomes more and more elongated.

## Turning the Definition of Ellipses on its Head

Often, this focal property is not thought of as a property of ellipses, but rather a defining feature. To see that these are equivalent, we have some to work to do. Let’s start by proving that stretched out circles actually have this focal property. We want to prove that for a “stretched out circle” defined by the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ and $b$ are two positive numbers, the sum of the distance between every point on the stretched circle and the two foci $\left ( \sqrt{a^2 - b^2},0 \right )$ and $\left ( - \sqrt{a^2 - b^2},0 \right )$ is the same. So we need to prove that for every point on the shape defined by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the sum of the distances to $\left ( \sqrt{a^2 - b^2},0 \right )$ and $\left ( - \sqrt{a^2 - b^2},0 \right )$ is the same number.

Proof: Suppose a point $(x,y)$ is on the curve defined by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Then we can solve for $y$ in the equation and express the point in terms of $x$:

$\frac{y^2}{b^2} & = 1 - \frac{x^2}{a^2} \\y^2 & = \frac{a^2b^2 - b^2x^2}{a^2} \\y & = \sqrt{\frac{a^2b^2-b^2x^2}{a^2}}$

Now, using the distance formula to compute the sum of the distance between the pairs of points: $\left (x, \sqrt{\frac{a^2b^2-b^2x^2}{a^2}} \right )$ and $\left ( \sqrt{a^2 - b^2},0 \right )$; and, $\left (x, \sqrt{\frac{a^2b^2-b^2x^2}{a^2}} \right )$ and $\left(-\sqrt{a^2-b^2},0\right)$ We have:

$\sqrt{\left (x-\sqrt{a^2 - b^2} \right )^2 + \frac{a^2b^2-b^2x^2}{a^2}} + \sqrt{\left (x+\sqrt{a^2 - b^2} \right )^2 + \frac{a^2b^2-b^2x^2}{a^2}}$

This algebraic equation looks daunting, but simplifying the expressions inside these square roots results in a surprising result. This becomes

$& \sqrt{\frac{a^2 \left (x - \sqrt{a^2 - b^2} \right )^2 + a^2b^2 - b^2 x^2}{a^2}} + \sqrt{\frac{a^2 \left (x+ \sqrt{a^2 - b^2} \right )^2 + a^2 b^2 - b^2x^2}{a^2}} \\& = \frac{1}{a} \left (\sqrt{a^2 \left (x- \sqrt{a^2 - b^2} \right )^2 + a^2b^2 - b^2x^2} + \sqrt{a^2 \left (x+ \sqrt{a^2 - b^2} \right )^2 + a^2b^2 - b^2x^2} \right ) \\& = \frac{1}{a} \left (\sqrt{a^2x^2 - 2a^2x \sqrt{a^2-b^2} + a^4 - a^2b^2 + a^2b^2 - b^2x^2} + \sqrt{a^2x^2 - 2a^2x\sqrt{a^2 - b^2} + a^4 - a^2b^2 + a^2b^2 - b^2x^2} \right ) \\& = \frac{1}{a} \left ( \sqrt{a^2x^2 - 2a^2 x \sqrt{a^2 - b^2} + a^4 - b^2x^2} + \sqrt{a^2x^2 - 2a^2x \sqrt{a^2 - b^2} + a^4 - b^2x^2} \right ) \\& = \frac{1}{a} \left (\sqrt{\left (a^2 - x \sqrt{a^2 - b^2} \right )^2} + \sqrt{\left (a^2 + x \sqrt{a^2 - b^2} \right )^2} \right ) \\& = \frac{1}{a} \left (a^2 - x \sqrt{a^2 - b^2} + a^2 + x \sqrt{a^2 - b^2} \right ) \\& = \frac{1}{a} (2a^2) \\& = 2a$

What a miraculous collapse! One of the gnarliest algebraic expressions that I have ever encountered turned into the simple expression $2a$. Most importantly, this reduced expression is merely a constant—it doesn’t depend on $x$ or $y$. This is exactly what we were hoping for. The sum of the distances between every point on the ellipse and the two foci is always $2a$. If you every find yourself needing to remember what this sum is, the number $2a$ can be remembered most easily by computing it from an easy point such as one of the shape’s $x-$intercepts, say $(a,0)$.

### Review Questions

1. Compute the distance from the $x-$intercept $(a,0)$ and the two foci and show that it is in fact $2a$.
2. What is the sum of the distances to the foci of the points on a vertically-oriented ellipse?

1. The distance between the $x-$intercept $(a,0)$ and $\left (\sqrt{a^2 - b^2}, 0 \right )$ is: $a - \sqrt{a^2 - b^2}$. The distance between the $x-$ intercept $(a,0)$ and $\left (- \sqrt{a^2 - b^2}, 0 \right )$ is: $a + \sqrt{a^2 - b^2}$. Together these add to: $a - \sqrt{a^2 - b^2} + a + \sqrt{a^2 - b^2} = 2a$
2. $2b$

## Defining an Ellipse by Focal Distance

So all this computing simply means that “stretched-out circles”—what we’ve been calling ellipses—satisfy the focal property. What would be great is if we could define ellipses by the focal property. This would be a nice generalization of the way we define circles.

Recall that circles are defined as the set of points in a plane that are a constant distance from a center point. Analogously, ellipses could be defined as a set of points in a plane for which the sum of the distances to two focus points is a constant. In other words, this definition would yield the exact same set of shapes as the “stretched out circle” definition that we started with.

Before proceeding we need to make sure of one thing. We’ve already proved that stretched out circles satisfy the focal property, but how do we know that any shape satisfying the focal property is in turn a stretched out circle? Well, the calculation above can simply be read backwards. In other words, suppose you have a set of points that satisfy the focal property, that each point whose sum of the distances to the points $(f,0)$ and $(-f,0)$ is a fixed distance $d$. Now note that for any two positive numbers $d$ and $f$ with $2f, there exist positive numbers $a$ and $b$ such that $d = 2a$ and $f = \sqrt{a^2 - b^2}$. (this fact is a bit subtle and is part of an exercise below.) So now we have a shape where every point $(x, y)$ has a sum of distances to the points $\left (\sqrt{a^2 - b^2}, 0 \right )$ and $\left (- \sqrt{a^2 - b^2}, 0 \right )$ which equals $2a$. Using these expressions, the algebraic steps of the proof above can simply be read backwards.

### Review Questions

1. Explain why for any two positive numbers $d$ and $f$ with $2f, there exist positive numbers $a$ and $b$ such that $d = 2a$ and $f = \sqrt{a^2 - b^2}$.
2. We just told you that the above proof could be read backwards. But you need to be careful when following algebraic steps backwards, especially ones involving squares or square roots.
1. For example, what happens when you follow this argument backwards? $x & = -2 \\x^2 & = 4$
2. Write a convincing argument that it is okay to follow the steps backwards in the above proof that every stretched circle has the focal property.

1. Set $a = \frac{d}{2}$. Then we need to show that for $f$ satisfying $2f, there exists a number $b$ such that $f = \sqrt{a^2 - b^2}$. Since $2f < d, 2f < 2a$ by the definition of $a$ (using the assumption $d > 0$). So $f < a$. We can find $b$ geometrically. Since $f < a$, there is a right triangle with one leg having length $f$ and hypotenuse $a$. Call the other leg of the triangle $b$. Then the Pythagorean Theorem tells us that $f^2 + b^2 = a^2$, or equivalently $f = \sqrt{a^2 - b^2}$. So $a$ and $b$ satisfy the necessary requirements.
1. Taking the square root of both sides of $x^2 = 4$ yields two solutions, $x = \pm 2$, instead of the one value we already know $(x = -2)$. The problem is that the operation of squaring a number is not a one-to-one function. Both $(-2)^2$ and $2^2$ yield the same number. So some information is lost during this step, and it cannot be perfectly “undone”, like other algebraic maneuvers.
2. The student’s reason should include the fact no information is lost (through squaring both sides or other operations) in any of these steps, so that each step is completely reversible.

## Equation of an Ellipse Not Centered at the Origin

All the ellipses we’ve looked at so far are centered around the origin (0,0). To find an equation for ellipses centered around another point, say $(h, k)$, simply replace $x$ with $x-h$ and $y$ with $y-k$. This will shift all the points of the ellipse to the right $h$ units (or left if $h < 0$) and to up $k$ units (or down if $k < 0$). So the general form for a horizontally- or vertically-oriented ellipse is:

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

It is centered about the point $(h,k)$. If $b, the ellipse is horizontally oriented and has foci $\left ( h + \sqrt{a^2 - b^2}, k \right )$ and $\left ( h - \sqrt{a^2 - b^2}, k \right )$ on its horizontal major axis. If $a, it is vertically oriented and has foci $\left ( h,k + \sqrt{a^2 - b^2} \right )$ and $\left ( h,k - \sqrt{a^2 - b^2} \right )$ on its vertical major axis.

Example 4

Graph the equation $4x^2 + 8x + 9y^2 - 36y + 4 = 0$.

Solution : We need to get the equation into the form of general equation above. The first step is to group all the $x$ terms and $y$ terms, factor our the leading coefficients of $x^2$ and $y^2$, and move the constants to the other side of the equation:

$4(x^2 + 2x) + 9(y^2 - 4y) = -4$

Now, we “complete the square” by adding the appropriate terms to the $x$ expressions and the $y$ expressions to make a perfect square. (See http://authors.ck12.org/wiki/index.php/Algebra_I-Chapter-10#Solving_Quadratic_Equations_by_Completing_the_Square for more on completing the square.)

$4(x^2 + 2x + 1)+ 9(y^2 - 4y + 4) = -4 + 4 + 36$

Now we factor and divide by the coefficients to get:

$\frac{(x+1)^2}{9} + \frac{(y-2)^2}{4} = 1$

And there we have it. Once it’s in this form, we see this is an ellipse is centered around the point (-1,2), it has a horizontal major axis of length 3 and a vertical minor axis of length 2, and from this we can make a sketch of the ellipse:

### Review Questions

1. Explain why subtracting $h$ from the $x-$term and $k$ from the $y-$term in the equation for an ellipse shifter the ellipse $h$ horizontally and $k$ vertically.
2. Graph this ellipse. $x^2 - 6x + 5y^2 - 10y - 66 = 0$
3. Now try this one that doesn’t have such nice numbers! $16x^2 - 48x + 125y^2 + 150y + 61 = 0$
4. Now try this one. $3x^2 - 12x + 5y^2 + 10y - 3 = 0$. What goes wrong? Explain what you think the graph of this equation might look like.
5. What about this one. $5x^2 - 15x - 2y^2 + 8y - 50 =0$? What goes wrong here? Explain what you think the graph of this equation might look like.

1. If $(x,y)$ is a solution to $\frac{(x-h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ then $(x+h,y+k)$ is a solution to $\frac{(x-h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$. This produces a graph that is shifted horizontally by $h$ and vertically by $k$.
2. After completing the square, we have the sum of positive numbers equaling a negative number. This is an impossibility, so the equation has no solutions.
3. After completing the square, the $x$ term and the $y$ term are opposite signs. If you plot some points you will see that the graph has two disconnected sections. This class of conic sections will be discussed in the next section.

## Difference Between an Ellipse and an Oval or Proving the Sliced Cone Definition of an Ellipse

There is still one critical step missing in our exploration of ellipses. We showed that “stretched-out circles” satisfy the focal property, and that any shape satisfying this property is in fact a “stretched-out circle”. So these are actually the same class of shapes, and they are called ellipses. But not any oval-shaped curve is an ellipse. Draw a random oval and you’re not likely to be able to find two points that satisfy the focal property. In particular, when we cut a cone with a tilted plane, how do we know that the oval-shaped curve that results is a “stretched out circle” satisfying the focal property?

Amazingly, the Ancient Greeks had an argument for this fact over two millennia ago. While it is impressive that this problem was solved so long ago, the argument itself involves an intricate construction that isn’t as illuminating as a more modern the one I’m going to show you instead. Most mathematicians prefer to use a more modern argument that is simply stunning in its simplicity. This modern argument isn’t fancy—the Greeks had all the tools they need to understand it—they just didn’t happen to think of it. It wasn’t until 1822 that the French mathematician Germinal Dandelin thought of this very clever construction. Dandelin found a way to find the foci and prove the focal property in one fell swoop. Here’s what he said.

Take the conic section in question. Then choose a sphere that is just the right size so that when it’s dropped into the conic, it touches the intersecting plane, as well as being snug against the cone on all sides. If you prefer, you can think of the sphere as a perfectly round balloon that is blown up until it “just fits” inside the cone, still touching the plane. Then do the same on the other side of the plane. After we’ve drawn both of these spheres we have this picture:

or

These spheres are often called “Dandelin spheres”, named after their discoverer. It turns out that not only is our shape an ellipse (which, like all ellipses satisfies the focal property), but these spheres touch the ellipse exactly at the two foci. To see this, consider this geometric argument.

The first thing to notice is that the circles $C_1$ and $C_2$ shown on the diagram below, where each sphere lies snug against the cone, lie in parallel planes to one another. In particular, each line passing through these circles and the vertex of the cone, such as the line $l$ drawn below, cuts off equal segments between the two circles. Let's call $d$ the shortest distance along the cone between circles $C_1$ and $C_2$. This can also be thought of as the shortest distance between $C_1$ and $C_2$ that passes through the vertex of the cone.

The next thing to remember is a property of tangents to spheres that you may have learned in geometry. If two segments are drawn between a point and a sphere, and if the line containing each segment is tangent to the sphere, then the two segments are equal. In the diagram below, $AB=AC$. (This follows from the fact that tangents are perpendicular to the radii of a sphere and that two congruent triangles are formed in this configuration. See this description for more about this property.)

Now consider the point $P$ on the ellipse drawn below. Let $\overline{QR}$ be the segment of length $d$ between $C_1$ and $C_2$ that passes through $P$. The distances between the two foci are marked $d_1$ and $d_2$. But $d_1 =RP$ and $d_2 =PQ$ by the property of tangents to spheres discussed above. So $d_1 + d_2 = RP + PQ = QR = d$. And this sum will always equal $d$, no matter what point $P$ on the ellipse is chosen. So this proves the focal property of ellipses: that the sum of the distances between any point on the ellipse and the two foci is constant.

### Review Questions

1. What do the Dandelin spheres look like in the case of a circle?
2. What is the area of an ellipse with the equation $\frac{(x-h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ ? (Hint: use a geometric argument starting with the area of a circle.)
3. What is the perimeter of an ellipse with the equation $\frac{(x-h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ ?

1. The Dandelin spheres for a circle lie directly above one another, and both touch the circle at the center point.
2. The area of an ellipse is $ab \pi$. To see why this is true, start with a circle of radius 1, which has an area of $\pi$. Then imagine an approximation with rectangles of the circle. Then stretch the rectangles by a factor of $a$ in the $x-$direction and by a factor of $b$ in the $y-$direction to obtain an approximation of the ellipse. This makes the rectangles $a$ times wider and $b$ times taller, giving an area that is $ab$ multiplied by the area of the approximation of the circle. Since this is true of any approximation of the circle, the area of the ellipse must be $ab \pi$.
3. This is actually a much more difficult question than the previous one. You’re on your own! Even the great Indian mathematician Ramanujan could only come up with an approximation: $p \approx \pi \left[3 (a+b) - \sqrt{(3a+b)(a+3b)}\right]$.

## Applications

The number of places ellipses appear in the natural world is immense. Consider, for instance, how often the simplest kind of ellipse, the circle, appears in your life. You see circles emanating as waves when you throw a rock into a pond. You see circular pupils when you look at a set of eyes. You see what is roughly a circle as the image of the sun or moon in the sky. The path of an object swung around on a string is a circle. When any of these circles are viewed at an angle you see an elongated circle, or an ellipse. So it is important to keep in mind that the discussion that follows covers only a few of the instances of ellipses in our daily lives.

## Planetary Motion

When a planet orbits the sun (or when any object orbits any other), it takes an elliptical path and the sun lies at one of the two foci of the ellipse. Johannes Kepler first proposed this at the beginning of the seventeenth century as one of three laws of planetary motions, after analyzing observational data of Tycho Brahe. His law is accurate enough to produce modern computations which are still used to predict the motion of artificial satellites. A century later, Newton’s law of gravity offers an explanation of why this law might be true.

### Review Questions

1. Though planets take an elliptical path around the sun, these ellipses often have a very low eccentricity, meaning they are close to being circles. The diagram above exaggerates the elliptical shape of a planet’s orbit. The Earth’s orbit has an eccentricity of 0.0167. Its minimum distance from the sun is 146 million km. What is its maximum distance from the sun? If the sun’s diameter is 1.4 million kilometers. Do both foci of the Earth’s orbit lie within the sun? Recall that the eccentricity of an ellipse is $\varepsilon = \frac{\sqrt{a^2 - b^2}}{a}$.
2. While the elliptical paths of planets are ellipses that are closely approximated by circles, comets and asteroids often have orbits that are ellipses with very high eccentricity. Halley’s comet has an eccentricity of 0.967, and comes within 54.6 million miles of the sun at its closest point, or “perihelion”. What is the furthest point it reaches from the sun?

1. Assume that the orbit of the sun is an ellipse centered at (0,0). Then we can use the distance from the origin to the focus $\sqrt{a^2 - b^2}$ to set up the equations $146 + 146 + 2 \sqrt{a^2 - b^2} = 2a$ and $0.167 = \frac{\sqrt{a^2 - b^2}}{a}$. Solving we get $a = 175.270, b = 175.245$, and the distance from (0,0) to the foci, $c = 2.927$ (all units are in millions of km). Finally the maximum distance from the earth to the sun is approximately 152 million km. From Kepler’s law, we know one of the foci of its orbit is at the center of the sun. The other foci is $2(2.927)=5.854$ million kilometers away, so it is outside the sun (but not by very far!).
2. $\sim 3.25$ billion miles.

## Echo Rooms

The National Statuary Hall in the United States Capital Building is an example of an ellipse-shaped room, sometimes called an “echo room”, which provide an interesting application to a property of ellipses. If a person whispers very quietly at one of the foci, the sound echoes in a way such that a person at the other focus can often hear them very clearly. Rumor has it that John Quincy Adams took advantage of this property to eavesdrop on conversations in this room.

The property of ellipses that makes echo rooms work is called the “optical property.” So why echoes, if this is an optical property? Well, light rays and sound waves bounce around in similar ways. In particular, they both bounce off walls at equal angles. In the diagram below, $\alpha = \beta$.

For a curved wall, they bounce at equal angles to the tangent line at that point:

So the “optical property” of ellipses is that lines between a point on the ellipse and the two foci form equal angles to the tangent at that point, or in other words, whispers coming from one foci bounce directly to the other foci. In the diagram below, for each $Q$ on the ellipse, $\angle{\alpha} \cong \angle{\beta}$.

This seems reasonable, given the symmetry of the ellipse, but how do we know it is true? First, let’s prove an important property that is a bit more general. Suppose you have two points, $P_1$ and $P_2$, and a line $l$, and you are interested in the shortest possible path between two $P_1$ and $P_2$, that intersects with $l$.

A nice way to find this is to reflect $P_2$ across $L$, obtaining the point $P_2'$ in the diagram below:

Since $l$ lies between $P_1$ and $P_2'$, it’s much easier to find the shortest distance between $P_1$ and $P_2'$ that passes through $l$. It’s simply the shortest path between $P_1$ and $P_2'$: a straight line! But for every path between $P_1$ and $P_2$, we can reflect the part from $l$ to $P_2$ to get a path equal in length between $P_1$ and $P_2'$. So the shortest path between $P_1$ and $P_2$ is simply the shortest path between $P_1$ and $P_2'$ (the straight line), with the part between $l$ and $P_2'$ reflected to get a path between $l$ and $P_2$. So the shortest path from $P_1$ to $P_2$ that intersects with $l$ is $P_1 Q$ followed by $Q P_2$.

The part of the above diagram that is going to help us prove the optical property is that since $\angle{1} \cong \angle{2}$ (vertical angles) and $\angle{2} \cong \angle{3}$ (reflected angles), then $\angle{1} \cong \angle{3}$ (transitive property). Thus the shortest path from $P_1$ to $P_2$ that intersects with $l$ consists of two segments that meet the line at equal angles.

Now to prove the optical property of the ellipse, apply the above situation to the ellipse. In the picture below, $P_1$ and $P_2$ are foci of the ellipse and $Q$ is the intersection with tangent line $l$.

The optical property states that $\angle{\alpha} \cong \angle{\beta}$. This is true because all points other than $Q$ on line $l$ lie outside the ellipse. Points outside the ellipse have a combined distance to the two foci that is greater than points on the ellipse. So $Q$ is the point on line $l$ with the smallest combined distance to points $P_1$ to $P_2$. Thus, like we showed in general above, $\angle{\alpha} \cong \angle{\beta}$. So the optical property has been proved.

### Review Questions

1. Design the largest possible echo room with the following constraints: You would like to spy on someone who will be 3 m from the tip of the ellipse. The room cannot be more than 100 m wide in any direction. How far from the person you’re spying on will you be standing.

1. The echo room has a major axis of 100 m and a minor axis of 34.12 m. Situating the room in the coordinate plane, the room can be represented by the equation: $\frac{x^2}{2500} + \frac{y^2}{291} = 1$. You will be 94 m from the person you are spying on.

## Sundials:

Conic sections help us solve the problem of making a sundial. Depending on the season, the sun shines at a different angle. However, due to the elliptical nature of the Earth’s orbit about the sun, a shadow-casting stick can be placed in such a way that the shadow always tells the correct time of day, no matter what the time of year, as long as the stick is lined up with the earth’s pole of rotation.

### Review Questions

1. No matter what the orientation of a stick, if you trace out the path that the shadow of the tip makes on a flat surface, you will find it is an ellipse. Describe why this is true.(HINT: for simplicity, you can assume that you are making the measurements throughout the course of one day and that with the exception of the earth rotating about a pole, the sun and the earth are fixed with respect to one another.)
2. It was mentioned earlier in the chapter that when a round glass of water is tilted, the surface of the water is an ellipse. Or, in other words, this statement is claiming that the cross section of a cylinder is an ellipse. Prove that this is true. (Hint: to prove that this is an ellipse all you need to do is show that for any cross section of a cylinder there exists a cone that has the same cross section.)
3. From the exercise above, it appears that there is some overlap between “conic sections” and “cylindrical sections”. Are any of the classes of conic sections we found in the last section not cylindrical sections? Are there any cylindrical sections that are not conic sections?

## Vocabulary

Ellipse
A conic section that can be equivalently defined as: 1) any finite conic section, 2) a circle which has been dilated (or “stretched”) in one direction, 3) the set of points in which the sum of distances to two special points called the foci is constant.
Major axis
The segment spanning an ellipse in the longest direction.
Minor axis
The segment spanning an ellipse in the shortest direction.
Focus
One of two points that defines an ellipse in the above definition.
Eccentricity
A measure of how “stretched out” an ellipse is. Formally, it is the distance between the two foci divided by the length of the major axis. The eccentricity ranges from 0 (a circle) to points close to 1, which are very elongated ellipses.

Feb 23, 2012

Dec 10, 2014