BQ #7: Unit V: Derivatives and the Area Problem

1. Explain in detail where the formula for the difference quotient comes from now that you know! Include all appropriate terminology (secant line, tangent line, h/delta x, etc).

 

 To start, let's look at this graph here:




There are two points where the red line (aka the secant line) and the blue curve cross. These two points are known as (x, f(x)) and (x+h, f(x+h)). h is not a point itself, but it represents the change in distance from point x to the second point on the x axis, which is why you sometimes see h notated as delta x (delta meaning 'the change in').

Sometimes, you'll also have points like this:



The purple line (aka the tangent line) is a line that only crosses the blue curve once. A graph like this has an infinite number of tangent lines, which have an infinite number of slopes at different points. But what if we wanted an equation that could give us the slope at any given point of the blue curve?

Let's go back to the secant line and its coordinates. We can easily find the slope of the secant line using the slope formula. Here it is in case you forgot it:



We plug in the two points (x, f(x)) and (x+h, f(x+h)) and get (f(x+h)-f(x))/(x+h)-(x)). The x's in the denominator cancel out, leaving us with (f(x+h)-f(x))/(h). Oh look, it's the difference quotient! It's also known as the derivative after you get far enough in math to start poking at calculus with a stick.



Why is this important? If you want to know the equation of the slope of the tangent line, you have to have h basically be 0. But it can't actually be zero because that would get you an undefined answer. So instead, you take the limit of the difference quotient of the blue curve as h approaches 0, plug in the x value of the coordinate where the tangent line and blue curve cross, and bam, you've got your slope. This can then be used to find all sorts of things, but that's for another time.

Resources:

www.jcu.edu

clas.sa.ucsb.edu

math.about.com

http://cis.stvincent.edu/carlsond/ma109/DifferenceQuotient_images/IMG0470.JPG 

BQ#6: Unit U: Limits, Functions, and the Like

1. What is a continuity? What is a discontinuity?

A continuous function is one that is predictable, has no breaks/jumps/holes, and is able to be drawn without lifting your pencil from the paper. In other words, you can predict where a continuous function is 'going' on the graph and there aren't any discontinuities (not to be confused with changes in equation) in the graph.

This is an example of a continuous function graph. As you can see, although it does change direction, the graph is still predictable after the change in equation, there aren't any discontinuities, and it is still able to be drawn without lifting pencil from paper.

A discontinuity is what makes a graph not continuous; it is a break, jump, or hole in the graph. There are four different types of discontinuities, separated into two families. The first family is the removable discontinuity family and it contains only point discontinuities, also known as holes.

This is a hole.

The second family is the non-removable discontinuity family, and it contains jump discontinuities, oscillating discontinuities, and infinite discontinuities.



The reason there are two families has to do with the limits (or lack thereof) of the discontinuities, which I will get to in the next question.

2. What is a limit? When does a limit exist? When does a limit does not exist? What is the difference between a limit and a value?

A limit is the intended height of the function- that is, it's where the limit wants to go. Sometimes it makes it there, and that's called a value. Continuous functions always have their limits and values equal each other. With removable discontinuities, it still reaches where it intends to go, but there isn't a value there; there's a hole. The limit of a point discontinuity doesn't match up with its value (which by the way is undefined), but there is still a limit. For non-removable discontinuities, the two sides of the function don't meet up at all and there is no limit. In this case, the limit Does Not Exist (or DNE for short) because the left side and right side of the function either approach different places (jump), approach so many places it becomes too wiggly to tell (oscillating), or has unbounded behavior caused by the presence of a vertical asymptote (infinite.)

3. How do we evaluate limits numerically, graphically, and algebraically?

 

To evaluate a limit numerically, we use a table like the one below.

 

 

We put what the limit approaches in the top center, go one tenth/hundredth/thousandth before and after it, and plug those numbers into the expression to find f(x) for each x value.

 

To evaluate a limit graphically, we use (big surprise) a graph. We look to see if the left and right sides of the function meet up. If they do, there is a limit (even if there's a hole.) If they don't, there is no limit and the reason depends on the discontinuity.

 

To evaluate a limit algebraically, we have three choices: direct substitution, dividing out/factoring, and rationalizing/conjugate.

 

The first method is direct substitution, which is exactly what it sounds like: taking whatever x approaches and plugging it into the expression. You should always use this method first because if you get a number over a number, a number over zero (aka undefined), or zero over a number (aka just plain zero), you're done! But if you get the dreaded zero over zero, that means there's a hole in the graph and you still need to do more work to figure out if the limit exists or not.

 

The second method is the dividing out/factoring method, which is also what it sounds like: looking to see if the numerator or the denominator has something to factor out and cancel. This removes the zero (and by extension, the hole, which is why point discontinuities are in the removable family.) After you've factored and canceled, plug in a la direct substitution and solve.

 

If direct substitution produces an indeterminate answer and factoring doesn't work, your last option is rationalizing/conjugate method. Multiply both the numerator and the denominator by the conjugate of the binomial term. The important thing to remember is NOT to factor the other term with the conjugate as the point of this is to get something to cancel, and it's easier to see if something is able to be canceled if you don't factor the other term with the conjugate. After you've FOILed the conjugates and canceled, use direct substitution and solve like you would normally.

 

Resources:

http://image.tutorvista.com/cms/images/39/graph-of-continuity-of-a-function.JPG

http://www.analyzemath.com/calculus/continuity/continuous_functions.html

http://web.cs.du.edu/~rjudd/calculus/calc1/notes/dis6.png

BQ#4: Unit T Concept 3: Upstair Downstairs...Graphs

Why is a “normal” tangent graph uphill, but a “normal” cotangent graph downhill? Use unit circle ratios to explain.

The parent tangent and cotangent graphs are in different directions because their asymptotes are in different places. According to the Ratio Identities, tan=sin/cos and cot=cos/sin. To get an asymptote, the ratio must be undefined (in other words, the denominator has to equal 0.)

How does all that fit in? Tangent has a denominator of cosine, so the places where cosine equals 0 on the Unit Circle and the tangent graph are 90º (pi/2) and 270º (3pi/2). However, cotangent has a denominator of sin, so the places where there would be an asymptote are 0º (0), 180º (pi), and 360º (2pi). Even though the pattern of positive and negatives are the same (+ - + -), because the asymptotes are in different places, the graphs have to be draw differently in order to not touch the asymptotes and follow the rules.

BQ#5: Unit T Concepts 1-3: What's Different About Sine and Cosine?

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do? Use unit circle ratios to explain.

Let's start by recapping what the Unit Circle ratios actually are, shall we?



If you will recall, x is the side adjacent to the angle, y is the side opposite to the angle, and r is the hypotenuse of the triangle. In the Unit Circle, r is always equal to 1. 

Now notice how sine and cosine are the only trig functions with a denominator of r. Since r always equals 1, sine and cosine will always be real numbers. For the other trig functions, there are places where the denominator in their ratio will equal 0. This makes the value of the trig function in question undefined, and this is where you get an asymptote. Since r will never equal 0, sine and cosine are never going to be undefined and therefore will also never have asymptotes.

Resources:

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh-VhUMaI7Gm3UZQdzH9R_WBfFHDd5OS4AKFI9EeCKWTyV6dgymbRguboI59U2nPnlUcgP5hRUcbJovdeBv5C4XjwaxwWhcEHnSS4a9NPEXcupf9LNTOVNPdXJOFZNKooaVZi9uTsclupXr/s1600/trigratios.png

BQ#3: Unit T Concepts 1-3: More Trig Graphs

How do the graphs of sine and cosine relate to each of the others?  Emphasize asymptotes in your response.
1.) Tangent?
Tangent is equal to sine/cosine, according to the Ratio Identity for tangent anyway. An asymptote is created wherever the denominator of a ratio equals 0 (and therefore is undefined), so whenever cosine's y value equals 0 on the graph, there will be an asymptote.



As you can see in this graph of cosine, the points where the y value is zero are -3pi/2, -pi/2, pi/2, and 3pi/2 (there are more, but let's not worry about that.)



In this graph of tangent, the points where the asymptotes are located are -3pi/2, -pi/2, pi/2, and 3pi/2 (again, there are more, but this is just a snapshot.)

2.) Cotangent?

Cotangent is equal to cosine/sine, also according to the Ratio Identity for Tangent. As stated before, an asymptote is found where the denominator equal 0. In this case, the denominator is sine, so the asymptotes are wherever sine's y value equals 0 on the graph.



As you can see in this graph of cosine, the points where the y value is zero are -2pi, -pi, 0, pi, and 2pi.



In this graph of cotangent, the points where the asymptotes are located are 0, 3.14, and 6.28- all of which are (approximately) the same points where sine equals 0.

3.) Secant?
Secant is equal to 1/cos, according to the Reciprocal Identity for secant. (We're done with Ratio Identities for now.) Secant has asymptotes too, which are located at the beginning and the end of each period.



Additionally, the minima and/or maxima of the secant graph also corresponds with the peaks and valleys of the cosine graph.

4.) Cosecant?

Cosecant is equal to 1/sin, according to the Reciprocal Identity for cosecant. Cosecant also has asymptotes (what a surprise), which are located at the beginning and the end of each period.



Additionally, the minima and/or maxima of the cosecant graph also corresponds with the peaks and valleys of the sine graph.

Resources:
http://www.mathsisfun.com/algebra/trig-sin-cos-tan-graphs.html
http://www.purplemath.com/modules/triggrph3.htm

BQ# 2: Unit T Concept Intro: Trig Graphs

How do the trig graphs relate to the Unit Circle?

A trig graph can be seen as an 'unwrapped' Unit Circle. The quadrants of a Unit Circle correspond with the sections on a trig graph. Whether or not the section of the graph is positive or negative -in other words, above or below the x axis- depends on the trig function graphed. For example, since sine is positive in Quadrants I and II and negative in Quadrants III and IV, the pattern of the sections of the trig graph for sine is + + - -, or above the x axis for two sections and below for the next two sections.

Period? - Why is the period for sine and cosine 2pi, whereas the period for tangent and cotangent is pi?
The period for sine and cosine is 2pi because it takes a length of 2pi along the x axis of the graph to complete the pattern of positives and negatives for sine and cosine. This length is also the length of the circumference of the Unit Circle (how coincidentally convenient!) The period for tangent and cotangent, on the other hand, is pi because it only takes a lengths of pi along the x axis of the graph to complete the pattern of positives and negatives for tangent and cotangent. Since tangent (and cotangent) is positive in Quadrant I, negative in Quadrant II, positive in Quadrant III, and negative in Quadrant IV, it only takes half of the circle to repeat the pattern.

Amplitude? – How does the fact that sine and cosine have amplitudes of one (and the other trig functions don’t have amplitudes) relate to what we know about the Unit Circle?
Sine and cosine have amplitudes because their waves are restricted. On the Unit Circle, sine and cosine cannot be bigger than 1, but neither can they be smaller than -1. The other trig functions don't have these restrictions in the Unit Circle, so they don't have restrictions in their graph either in the form of amplitudes.

Reflection #1: Unit Q: Verifying Trig Identities

1.) What does it actually mean to verify a trig identity?

To verify a trig identity means to simplify an expression with trig functions until it equals something. When someone asks you to 'verify a trig identity', they mean 'look, here's two expressions that are set equal to each other. Make one side look like the other side.' Usually this consists of simplifying the more complex side to that it has the same terms as the simpler side. It also usually involves substituting trig identities to get things to cancel (and therefore simplify.)

2.) What tips and tricks have you found helpful?

I personally like to convert everything to sine and cosine. It makes it easier for me to see what cancels and what can be substituted (particularly with the Pythagorean Identities.) I also find it helpful to use the Reciprocal Identities to convert everything to the same trig function. Again, it makes it easier for me to see what cancels. Another tip is to try to get things to cancel to 1 (when there's multiplication or division) or 0 (when there's addition or subtraction.)

3.) Explain your thought process and steps you take in verifying a trig identity.Do not use a specific example, but speak in general terms of what you would do no matter what they give you.

Like I explained in the second answer, I like to try to use the Reciprocal Identities to cancel things out. I don't like to convert everything to sine and cosine until later on in the solving process. This is because converting to sine and cosine too early in the solving process can make things needlessly complicated. My last resorts are squaring the expression or multiplying by a conjugate (because those are almost guaranteed to make things very complicated.)

SP#7: Unit Q Concept 2: Pythagorean/Ratio/Reciprocal Identities with SOH CAH TOA

This SP was made in collaboration with Leslie N. and can be seen by clicking this word right here. While you're there, be sure to check out her awesome math posts too!

ID# 3: Unit Q Concept 1: Pythagorean Identities

Inquiry Activity Summary

1.) Where does where sin2x+cos2x=1 come from to begin with (think Unit Circle!).

Well to answer this question we first have to think about the Pythagorean Theorem. It's an identity (which means it's a proven formula hat's always true) and we used it a lot when dealing with the Unit Circle. The variables we used were x (later we'll use x in the final form but in that case x stands for the angle), y, and r; so the formula was x^2 + y^2 = r^2. (I'll explain why the Unit Circle is important in a bit.)

 

Let's say we wanted the equation to equal 1. The easiest way to make that happen is to divide everything by r^2, which would leave us with x^2/r^2 + y^2/r^2 = 1.

 

Coincidentally, the cosine and sine ratios for the Unit Circle are x/r and y/r. Huh. Scratch that, not so coincidental.

 

If we replaced the ratios with sine and cosine (and switched the positions of sine and cosine but that's okay because it's addition), we'd get sin^2x + cos^2x = 1, which is the Pythagorean Identity!

 

2.) Show and explain how to derive the two remaining Pythagorean Identities from sin2x+cos2x=1.  Be sure to show step by step.
 

The remaining two Pythagorean Identities are 1 + tan^2x = sec^2x and 1 + cot^2x = csc^2x.

 

We'll start with the Pythagorean Theorem again: x^2 + y^2 = r^2. Instead of dividing by r^2, let's divide by x^2 and make the x term equal to 1. When we do that, we end up with  1 + y^2/x^2 = r^2/x^2. In the Unit Circle, y/x is the ratio for tangent and r/x is the ratio for secant. Replace them and you end up with 1 + tan^2x = sec^2x- another Pythagorean Identity!

 

Two down, we've got one to go. Start with the Pythagorean Theorem again: x^2 + y^2 = r^2. We've already used r^2 and x^2, so let's go with our last variable squared y^2. When you divide everything by y^2, you get x^2/y^2 + 1 = r^2/y^2. In the Unit Circle, x/y is the ratio for cotangent and r/y is the ratio for cosecant. Replace them (and switch the x and y terms around if you like) and you end up with 1 + cot^2x = csc^2x- the last Pythagorean Identity.

 

Inquiry Activity Reflection

1.) “The connections that I see between Units N, O, P, and Q so far are…”

 

The connections I see between the three units are the use of the Unit Circle and the use of trigonometric ratios.

 

“If I had to describe trigonometry in THREE words, they would be…”
 

The three words would be "memorization", "ratios", and "angles".

WPP#13 &14: Unit P Concepts 6 & 7: Laws of Sine and Cosine

This WPP was a collaboration between myself and Leslie N. Go here to check out more of her cute math posts! If you'd rather stay here, luckily for you there's a lovely WPP right below for you to check out her math anyway~

Create your own Playlist on LessonPaths!

BQ# 1: Unit P: Concepts 1 and 4

[1] Law of Sines
Why do we need it?
We need the Law of Sines when we are not sure if a triangle is a right triangle.

How is it derived from what we already know?
Since we don't know if the triangle we're given is a right triangle, we can't use the trig ratios we used in the Unit Circle. However, we can use trig functions. Say we're given the triangle below.



We don't know if it is a right triangle, but that doesn't matter because we can just do this

and bam! We've got two right triangles! We can call the green perpendicular line h.

If you remember geometry trig, you'll remember a handy mnemonic: SOH CAH TOA. SOH means "sine equals opposite over hypotenuse", and in this case, the opposite side from angle A is the green line h while the hypotenuse is side C. Therefore, we can say that sineA= h/c.

But what about the other triangle we made? We can apply SOH to this triangle too, and we get sinC=h/a.

As you can see, h is in both of these equations, so if we fiddle a bit to get h by itself in both equations (multiplying both sides by c in the first, multiplying both sides by a in the second), we can set the two equal to each other. Now we have c(sinA)= a(sinC). Fiddle around a bit more (and by that I mean divide both sides by ac- a term both sides have) and you're left with the Law of Sines...sort of.



It's almost there, but what about angle B? We can't just leave it all by itself!

 Not to worry, we can just make another green perpendicular line and do the same thing, only with different angles.

In this case, SOH for the two triangles are sinA=h/b and sinB=h/a. Do the same fiddling to get h by itself to set the two equations equal to each other and we get b(sinA)=a(sinB). Fiddle again and we end with this lovely bit:

Still not the Law of Sines, but we're close. As you can see, both equations have sinA/a in them. Which means...we can set them equal to each other! Yay! And for added fun, we don't even have to get rid of sinA/a! We can just mash them together to get this:



And there's the Law of Sines. Isn't it lovely?

[4] Area Formulas
How is the “area of an oblique” triangle derived?  How does it relate to the area formula that you are familiar with?I'm going to go ahead and answer both of these questions in one chunk. Less reading, how lucky for you!

Below are some examples of an oblique triangle (aka a triangle with all sides not being equal.)



As you can see, we start by doing the same thing we did before- make right triangles with a perpendicular line called h because mathematicians love right triangles more and we already know how to work with them.

If you recall from reading the Law of Sines section above, you'll know that we can get h by itself at some point with any of the angles. you have the option of a(sinB), a(sinC), b(sinA), b(sinC), c(sinA) and c(sinB) depending on which angle you sacrificed to draw h. Why is getting h by itself so important, you might ask? Again, if you remember geometry, you'll know the area of a triangle is usually given as a=1/2(b)(h). Oh look at that, there's an h in that equation that we can conveniently substitute any of the above options into, how fortunate for us!

Let's go with the first option, a(sinC) for simplicity's sake. When we substitute that value into the area equation, we get...

/drumrolls/

...1/2 (b)(a)(sinC)! And that's our area of an oblique triangle equation.

References:
http://www.mathsisfun.com/algebra/trig-sine-law.html
http://www.regentsprep.org/Regents/math/algtrig/ATT12/lawofsines.htm
http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

WPP#12: Unit O Concept 12: Angles of Elevation and Depression

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I/D# 2: Unit O Concepts 7-8: Deriving Patterns for the SRTs

Inquiry Activity Summary

1.) 30-60-90 SRT

We start with an equilateral triangle with side lengths 1 and split it in half in a way that bisects an angle and is perpendicular to the side opposite.

Each angle has a value of 60º (so bisecting them would get two 30º angles) and each side has a value of 1 (so making the 'cut' perpendicular to the side opposite would get you two 90º angles and side lengths of 1/2.)

Use the Pythagorean theorem (a^2 + b^2 = c^2) to find the missing leg length. 1/2 (let's say it's a) squared is 1/4, and 1 (let's say it's c) squared is just 1. When you subtract 1/4 from 1 to solve for b you get 3/4, the square root of which is radical 3 over 2.

To get rid of the fractions and make life easier for ourselves we multiply all the terms by 2. This results in the shortest leg equaling 1, the other leg equaling radical 3, and the hypotenuse equaling 2. We now set n equal to 1 (we choose n because it's a variable and we choose 1 because it's in all of the terms) so we can use it for number values other than 1, 2 and radical 3. Therefore, the pattern is n (shortest leg), 2n (hypotenuse- think of it as 2 times 1 and replace the 1 with n), and n radical 3 (other leg- again, think of it as 1 times radical 3 and replace the 1 with n.)

2.) 45-45-90 SRT

We start with a square with side lengths of 1 and split it in half diagonally.

This bisects the corner angles, which have a value of 90º each. 90/2 = 45º.

We know the two legs of the triangle each equal 1, so to find the hypotenuse we use the Pythagorean theorem (a^2 + b^2 = c^2). a and b represent the legs of the triangle, so we set them both equal to 1; 1 squared is just 1, so 1 + 1 = 2. c therefore equals radical 2.

We now set n equal to 1 (we choose n because it's a variable and we choose 1 because it's in all of the terms) so we can use it for number values other than 1 and radical 2. Therefore, the pattern is n (one leg), n (other leg), and n radical 2 (for the hypotenuse- think of it as 1 times radical 2 and replace the 1 with an n.)

Inquiry Activity Reflection

1.) “Something I never noticed before about special right triangles is…” that they were derived from an equilateral triangle and a square.
2.) “Being able to derive these patterns myself aids in my learning because…” I now know how to derive the side lengths myself instead of relying on my subpar memory.

I/D# 1: Unit N: How Do Special Right Triangles and the Unit Circle Relate?

Inquiry Summary Activity

1.) This is a 30º triangle.

As you can see, the properties of a 30º SRT are that the side opposite the angle has a value of x, the side adjacent to the angle has a value of x radical 3, and the hypotenuse has a value of 2x.

 

We simplify the hypotenuse to make it equal to 1, but whatever we do to one side must be done to the other sides. When we divide the hypotenuse by 2x, we also divide the opposite and adjacent sides by 2x, which reduce to 1/2 and radical 3 over 2 respectively.

 

 

 

When we place the triangle on a coordinate plane so that the angle is at the origin, we can figure out the coordinates of the points of the triangle. The point with the 30º angle is at (0,0), the point with the right angle is at (radical 3 over 2, 0), and the last point is at (radical 3 over 2, 1/2).

 

2.) This is a 45º triangle.

 

As you can see, the properties of a 45º SRT are that the hypotenuse has a value of x radical 2 and the other two sides have a value of x.

 

We simplify the hypotenuse to make it equal to 1, but whatever we do to one side must be done to the other sides. When we divide the hypotenuse by x radical 2, we also divide the other sides by x radical 2, which both reduce to radical 2 over two.

 

 

 

When we place the triangle on a coordinate plane so that the angle is at the origin, we can figure out the coordinates of the points of the triangle. The point with the 45º angle is at (0,0), the point with the right angle is at (radical 2 over 2, 0), and the last point is at (radical 2 over 2, radical 2 over 2).

 

3.) This is a 60º triangle.

As you can see, the properties of a 60º SRT are that the side opposite the angle has a value of x radical 3, the side adjacent to the angle has a value of x, and the hypotenuse has a value of 2x.

 

 We simplify the hypotenuse to make it equal to 1, but whatever we do to one side must be done to the other sides. When we divide the hypotenuse by 2x, we also divide the opposite and adjacent sides by 2x, which reduce to radical 3 over 2 and 1/2 respectively.

 

 

When we place the triangle on a coordinate plane so that the angle is at the origin, we can figure out the coordinates of the points of the triangle. The point with the 60º angle is at (0,0), the point with the right angle is at (1/2, 0), and the last point is at (1/2, radical 3 over 2).

 

4.) This activity helps us derive the Unit Circle by showing us how to get the ordered pairs that make up the circle. When we make the hypotenuse equal to 1, we essentially set it as the radius of the circle. We also now understand which ordered pairs are connected with which SRTs.

 

5.) For this next section, we will be focusing specifically on the coordinates of the radius- or rather, how they change when the SRTs are moved to different quadrants.

 

 

When we move the 30º triangle to the second quadrant, the y value of the ordered pair stays the same, but the x value is now negative because it is on the negative side of the x axis.

 

 

When we move the 45º triangle to the second quadrant, the x value is still negative because it is on the negative side of the x axis, and the y value is now also negative because it is on the negative side of the y axis as well.

 

When we move the 60º triangle to the second quadrant, the x value is positive again because it is back on the positive side of the x axis, and the y value is still negative because it is still on the negative side of the y axis.

 

Inquiry Activity Reflection

 

1.)  “The coolest thing I learned from this activity was" how SRTs can be used to find the 'Magic 3' ordered pairs of a Unit Circle.

 

2.)  “This activity will help me in this unit because…” I now know how to derive the ordered pairs of the UC, so now I don't have to rely on my subpar memory when drawing it out.

 

3.) “Something I never realized before about special right triangles and the unit circle is…” the properties of the SRTs actually play a big part in deriving the ordered pairs of the UC.

RWA# 1: Unit M Concepts 4-6: Conic Sections in Real Life

1. Definition
An ellipse is the set of all points such that the sum of two points called foci equals a constant.

2. Description
The equation of an ellipse is as follows:

(x-h)^2 + (y-k)^2

   a^2          b^2

The x and y variables represent the coordinates of a point on the graph of the ellipse. The h and k variables represent the x value and the y value of the center, respectively. The a and b values determine the distance of the side of the ellipse from the center along the major and minor axes, respectively. The length of the major axis is double the value of a, and the length of the minor axis is double the value of b. To differentiate between a and b, you must keep in mind that a will always be the larger number, regardless of whether it is under the y term or the x term.

The graph of an ellipse is as follows:

http://www.csgnetwork.com/ellipse.png

As you can see, there are more variables that cannot be known just by looking at the equation. The c value is the distance from the center to the foci. It is always located along the major axis. To find c, use the equation a^2 - b^2 = c^2. The foci are two points that are also along the major axis. When you connect the foci to another point on the ellipse (see the blue), the sum will always be a constant. Additionally, the closer the foci are to the center, the more circular the ellipse will be. This value of how much the conic section deviates from being circular is called eccentricity. To find eccentricity, use the equation c/a.

3. Real World Application
One special property of an ellipse is that energy or sound directed at one focus will reflect to the other focus. A real world application of this property is called a whispering gallery. One person can stand at the focus of the whispering gallery and say something. However, whatever they say will not be heard by anyone except someone who is at the other focus. The sound waves bounce off the sides of the ellipse and reflect to the other focus.

A video of how the foci of an ellipse are related is as follows:

As the video shows, the sum of the distance between the two foci and the two points stays constant throughout the formation of the ellipse. A sound wave travelling a certain distance through one of the foci and bouncing off the side of the wall is like making the first segment. Since it can only go a certain distance more, it must reflect toward and through the other focus.

4. Works Cited
Image of an ellipse found here.
Definition of a whispering gallery found here.
Video link for the construction of an ellipse found here.