WPP#13/14: Unit P-Concepts 6 and 7

Photo-credits:

http://24.media.tumblr.com/48a60113d127af3168cd9ff610d86684/tumblr_my0biwlZRg1qmxwggo1_r2_500.jpg

Hazel grace is located 135 miles due south of Augustus Waters. Destiny is calling out for their first meet in order for them to meet and fall in love. Destiny's call indicates that Hazel Grace is located at a bearing of 60 degrees; Destiny's call indicates that Augustus Waters is located at a bearing of 131 degrees. How far are Hazel Grace and Augustus from Destiny?

Photo-credit: lovely, hand-drawn pictures by Mrs. Ventura

Two planes leave the airport at the same time, traveling on courses that have an angle of 125 degrees between them. If the first plane travels at 35 miles per hour and the second travels 28 miles per hour, how apart are the two planes after traveling for 5 hours?

Photo-credit: lovely, hand-drawn pictures by Mrs. Ventura

BQ#1- Concepts 1 and 4

1. Law of sines-Why do we need it? How is it derived from what we already know?

Photo-credits: lovely, hand-drawn pictures by Mrs. Ventura

Using the picture from above, we will go step-by-step of the derivation of the equation. First things first, we must know that the law of sines is needed to solve missing parts of a non-right triangle. A regular right triangle can be easily solved because it has all the needed components, that being, most importantly, a right triangle. So, splitting a non-right triangle we get two right triangles. We solve for both angles A and C using sin. From there we get sinA=h/c and sinC=h/a. We take it one step further and we multiply by c  and a ,for the two separate equations, in order to get h alone. Having h alone for both equations, we then combine them and move on over the a and c to their appropriate sides. From there we get the law of sines. Of course the law of sines is not just restricted to a or c, it also works for b as well. The law of sines fluctuates according to what specific information you are looking for.

Photo-credits: lovely, hand-drawn pictures by Mrs. Ventura

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?

The area of an oblique triangle is derived from the area formula and a combination of a trig function. The area formula is A= 1/2bh. With a non-right triangle, trying to solve the area is much more difficult. Therefore, we split the triangle with a perpendicular line and get two triangles. From there, we find angle C. SinC=h/a, we multiply a to the other side in order to get h alone. Moreover, we plug in h into the original area formula. That, is the equation for solving the area of an oblique triangle.

It's similar to the area formula we know, except we manipulate it, in order to get what we need. The only difference is since we do not have h we use our sources and use it to the best of our abilities.

Below are other equations that are used to solve oblique triangles:

Photo-credits: lovely, hand-drawn pictures by Mrs. Ventura

BQ #7- Where does the difference quotient come from?

Photo-credits: lovely, hand-drawn pictures by Victoria

Looking above at the image, we notice that our first points the x-value is x and our y-value is f(x). Technically looking at the graph not a certain distance is given so that's the reason why we use f(x). For the very same reason the next set of points x-value is x+h and the y-value is f(x+h). The sole purpose of this graph is to find the slope tangent line on the two specific points. To find the slope of the tangent line we must use the slope formula which is m=(y2-y1)/(x2-x1). We plug in the points to the formula and continue on simplifying as much as we can: cancelling/combining like terms, simplifying, etc. Once we have finished the whole lovely process we get the difference quotient! Voila! The difference quotient is extremely helpful in finding derivatives for a certain graph. A visual is displayed below for the process of plugging in to the points to the slope formula ---> to becoming the difference quotient.

Photo-credits: lovely, hand-drawn pictures by Victoria

BQ#6: Unit U-CALCULUS INTRO WOOO

1. What is continuity? What is discontinuity?

 

Photocredits: http://www.mathsisfun.com/calculus/continuity.html

In order for a graph to be continuous, first of all, must be predictable. Secondly must not have any jumps, holes or vertical asymptotes. Lastly, must be drawn without lifting the pencil off the paper. All of these elements can be seen in the pictures above. Now discontinuities are the exact opposite of this. But do note that there are two different types of discontinuities: removable and non-removable. A question frequently asked is why are there two groups and that my friends is because removable are when limits do exist and non-removable discontinuities do not exist. One removable discontinuity is point and the limit here does exist because although it does not reach the intended height it does reach a certain height the actual height and that is seen through the value. The three non-removable discontinuities are jump, oscillating and infinite. A limit cannot exist in jump because there are differences in left and right graphs and they do not meet. A limit does not exist at oscillating because, well, it cannot be found. The graph is too out of control to actually find where the limit exists. In an infinite discontinuity a limit does not exist due to unbounded behavior.

Photocredits: Unit U SSS Packet by Mrs. Kirch

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

A limit is the intended height of a function. A limit exists when a graph is continuous or as previously stated when, no jumps, holes or vertical asymptotes are present. A limit does not exist when there are left and right differences (jump), when the graph is wiggly (oscillating) and when there is unbounded behavior (infinite). All of these stated are elements of non-removable discontinuities. Lastly, there is a BIG difference between a limit and value. A limit is the intended height of a function, BUT the value is the actual height of the function. When solving equations it is extremely important to know the difference between the two knowing it will affect your answer (ya know, whether you get it right or wrong. OBVI you wanna get it right soo...know the difference!).

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

Numerically- To solve limits numerically a table would be necessary and key to help you solve. The limit statement that is said verbally. Is "the limit as x approaches # of (blank) is #. Or as seen in the image below

Photocredits: Unit U SSS Packet by Mrs. Kirch

Graphically-

Photocredits: Unit U SSS Packet by Mrs. Kirch

As seen above a limit was solved graphically. It is vital to know the difference of the value and limit here because solving it graphically makes it more difficult. I through M are all limits for I and M since they do not exist a b/c is necessary in order to fully understand why the limit DNE. Sometimes, because a limit does not exist, it does not mean that a value is present. So when solving a limit graphically it is HIGHLY important to pay attention to all the little details, they are very important.

Algebraically- There are three ways to solve a limit algebraically those being: direct substitution method, dividing out/factoring method, and rationalizing/conjugate method.

FIRST method to try out would be the substitution method in which you substitute all the variables with the give number that x approaches in the limit statement. If it turns out to be undefined the next step to trying to solve it out is using the dividing out/factoring method. In this method we factor out any long equations and then find common factors and divide them out. With what is left from there we substitute the given number x approaches in the limit statement given and if we get an answer AWESOME-SAUCE and if we don't we go for our last and final method that being: the rationalizing/conjugate method! In this method you multiply by the conjugate in the denominator to eliminate some things and make life much easier, this method is the ONE. And most of the time works out fine so whenever the other two methods do not make your life easier, come to this one! If more help is needed in understanding the last method, look at the image below:

Photocredits: Unit U SSS Packet by Mrs. Kirch

BQ#4: Unit T Concepts 1-3-Why is a "normal" tangent graph uphill, but a "normal" cotangent graph downhill?

Photocredits: desmos.com

Yes, tangent and cotangent are inverses of each other but, that does not fully explain why exactly a tangent graph is uphill and a cotangent graph is downhill. Let's start off with tangent first and why it's uphill. Tan=sin/cos in order for tan to be undefined and have asymptotes cos must be zero in certain points of the unit circle: pi/2 (90degrees) and 3pi/2 (270degrees). Co-tan, unlike tan, has asymptotes in other points on the unit circle, it's asymptotes are on: 0pi, pi (180 degrees), 2pi (360 degrees). And that completely answers our question. Tangent graphs and cotangent graphs go in two different directions because of the different asymptotes, they are undefined in different areas, so, therefore, they go in different directions.

Photocredits: desmos.com

BQ#3:Unit T-How do graphs of sine and cosine relate to each of the others?

a. Tangent?

                                                         Photocredits: desmos.com

In order to fully understand how all trig functions seen on this graph relate, lets go quadrant by quadrant. In the first quadrant, sin, cos and tan are positive. So, therefore, all of the graphs at this point are above the x-axis showing they are positive. In the second quadrant, it's quite different sin is the only trig function that's positive in this quadrant the other two are negative. In quadrant three, tan is the only positive trig and the other two are negative. Lastly, in quadrant four, cos is the only positive and the other two are negative.

So, what's the big deal with these asymptotes and why are they placed in the places they are placed? Tangent has asymptotes because at certain points such as pi/2(90 degrees) and 3pi/2 (270 degrees) tan's ratio (x/y) is undefined. That is why the tangent graph looks split in between sections, the tangent periods don't touch the asymptotes but do get very close. 

b. Co-tangent?

                                                                                   Photocredits: desmos.com

Cotangent's graph is very similar to tangent, positive at all the same parts and negative at all the same parts, except that they go in two different directions. Why? Well, because yes they're inverses of each other but they have asymptotes at different points on the unit circle. In the Unit Circle tan's ratio is undefined at 0, pi (180 degrees) and 2pi (360 degrees). Therefore, the co-tangent graph goes downhill.

c. Secant?

                                                                                   Photocredits: desmos.com

Secant, the beautiful secant. So, what's up with this so-called secant? In the first quadrant, all three, sin, cos and secant are positive. In the second quadrant, sin is the only one positive, the other two are negative. Furthermore, the third quadrant, all (sin, cosine and secant) are negative in this quadrant. Lastly, in the fourth quadrant, both cos and secant are positive, and the only one negative is sin.

The asymptotes for secant are on pi/2(90 degrees) and 3pi/2 (270 degrees). Those are the specific asymptotes for secant because that is where secant's ratio is undefined.

What seems so odd is that the asymptote of secant is connected with cos, why is that? That is because in simple words secant is the inverse of cosine! 

d. co-secant?

                                 Photocredits: desmos.com

Co-secant is the inverse of sine. In the first quadrant, all three, sin, cosine and co-secant are positive. In the second quadrant, sin and co-secant are positive and cos is negative. Moreover, in the third quadrant, all the three trig functions are negative. And lastly, in the fourth quadrant cos is the only one positive, the other two (sin, co-secant) are negative.

Much like co-tangent, co-secant has certain asymptotes because that is where its ratio is undefined. Co-secant is undefined at at 0, pi (180 degrees) and 2pi (360 degrees).

And just like secant, co-secant is attached to sin because its the inverse of co-secant!

Thank you for reading my blog. Comments, concerns, questions? Feel free to call me *wink*. Just kidding, comment below..

BQ#5: Unit T Concepts1-3-Why do sine and cosine NOT have asymptotes, but the other four trig graphs do?

First question that should be answered is: how can you get an asymptote? When a trig function is undefined is where an asymptote would appear on a graph. (Okay, got that cleared.) We must know that in the Unit Circle r=1. So therefore, sine and cosine can never be undefined because they would always have a one underneath in their ratio. Unlike sine and cosine, co-secant, secant, tangent and cotangent CAN have asymptotes. Why? Well for one, they do not have "r" underneath as their denominator in their ratio. Meaning, they can possibly have one of their values on the bottom as 0. For instance with secant, its ratio is r/x, we know that "r" is 1 and possibly if "x" is 0, our answer would be undefined. For co-secant, secant, tangent, and cotangent there are many possibilities that an answer can be undefined and that is why these trig functions have asymptotes and, sine and cosine do not. Sine and cosine will never be undefined nor have an asymptote. 

ANY QUESTIONS? FEEL FREE TO COMMENT BELOW, THANK YOU.

BQ#2: Unit T Intro-How do the trig graphs relate to the Unit Circle?

Terms to help for the reading below:

         PERIOD-graphs are cyclical and once completion of one cycle its called a period.

         AMPLITUDES-are half the distance between the highest and lowest points on the graph.

Photo-credit: Fabulous and Beautiful Victoria Ventura

Photo-credit: Fabulous and Beautiful Victoria Ventura

A. Period?-Why is the period for sine and cosine 2pi, whereas the period for tangent and cotangent is pi? Interesting question. As seen from the lovely drawn pictures from above, we notice that both in sine and cosine we have repeating patterns: +,+,-,- and +, -, -,+. In order to complete a period we must go all around the unit circle. Keep in mind our graphs are just the unit circle unfolded and repeated with the same period. From 0 degrees to 360 degrees is 2pi. So that is why sine and cosine are 2pi. Seen from the picture below, we see that the pattern is +,-,+,-. The reason why tangent and cotangent's period is only pi is because the pattern is already completed by pi, unlike sine and cosine, in which one has to go all around in order to complete the period. Life is much easier doing tangent and cotangent, to be honest. 

Photo-credit: Fabulous and Beautiful Victoria Ventura

B. 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? In the Unit Circle all the other trig functions could be greater than one (YAY, for them) but for sine and cosine, they couldn't be greater than one. Which makes complete sense why the amplitude for sine and cosine is restricted to one and the others CAN be greater than one (such rebels.)


ANY QUESTIONS FEEL FREE TO COMMENT BELOW, THANK YOU FOR YOUR TIME.

Reflection #1: Unit Q Verifying Trig Identities

1. What does it actually mean to verify a trig identity?
What verifying a trig function actually means is to prove that the answer given can actually be solved out from the equation given. There are multiple ways to verify a trig function. It might, at first, seem easy but because we are given an answer, we have to make sure to get the exact answer. There are multiple ways to verifying a trig identity but one must be cautious in what other trig identity is being used.

2. What tips and tricks have you found helpful?
Definitely memorizing all trig identities is very helpful. Using reciprocal identities come very handy when solving. If you see a sin, cos or tan squared, figure out if you can use a Pythagorean identity! It will honestly save your life. Sometimes ratio identities can save your life too. Although it is important to know the identities off the top of your head, while doing pq's, or homework it is also helpful to have a list of the identities out in front of you.

3. Explain your thought process and steps you take in verifying a trig identity.
First thing is DO NOT FREAK OUT. I freak out all the time and instead of doing the identity I stare at it and how hard it looks instead of doing it! Then I figure out I can use the reciprocal identities, cross multiply and BOOM, I get the answer. A trig identity might look extremely complex but all it might take is replacing one of the parts with a Pythagorean identity, another with a reciprocal identity and that can lead to your answer. Remember, do not freak out, be calm and solve your life away!

SP#7: Unit Q Concept 2-Finding trig functions using identities

This post was made in collaboration with Jorge Molina. Please visit the other awesome posts on his blog by going here.

Hello there fellow students. Before starting the problem there are a few things you must be aware about. First of all, make sure not to solve this problem straight using trigonometry. We will be using identities for this problem, including: reciprocal identities, ratio identities, and Pythagorean identities. Secondly, solving identities are much easier than we think of them to be. Make sure to look at the clues to help you solve for the problem. Lastly, make sure when rationalizing to be careful what you are multiplying and always look for simplification. Other than that, relax, enjoy and solve your life away!

Photo-credits to the fab-u-lous Victoria Ventura

Photo-credits to the fab-u-lous Victoria Ventura

Photo-credits to the fab-u-lous Victoria Ventura

Photo-credits to the fab-u-lous Jorge Molina

Photo-credits to the fab-u-lous Jorge Molina

Photo-credits to the fab-u-lous Jorge Molina

Photo-credits to the fab-u-lous Victoria Ventura

WPP#12: Unit O Concept 10-Solving angle of elevation and depression word problems

Pooky's great adventure in Disneyland!:

One beautiful Tuesday afternoon, while roaming around looking for her parents, Pooky stumbled upon the Sleeping Beauty Castle at Disneyland. She was admiring its beauty when all of the sudden she was kidnapped by the troublesome Tinkerbell! GASP. (Here's the deets: Tinkerbell was jealous that as Peter Pan flew past the castle, he admiringly stared at Pooky.) Unfortunately, Tinkerbell placed Pooky in the highest tower, making it difficult for Pooky to be rescued.

While Pooky waited for Prince Charming to save her from the tower, she decided to measure Matterhorn's mountain height. Being the smart-alec she was, she figured out something incredible.

She measured the angle of elevation to the mountain across to be 29 degrees and the angle of depression (to the base of the mountain) to be 40 degrees. If the two, the castle and mountain, are 75 feet apart, how tall is the Matterhorn Mountain? (Round to the nearest foot) 

Here's a lovely picture of Pooky admiring Disneyland:

Picture made and taken by the lovely Victoria Ventura

Here's the real business and work for the problem:

Picture made and taken by the lovely Victoria Ventura

ANY QUESTIONS? FEEL FREE TO COMMENT.

I/D#2: Unit O-How can we derive the patterns for our special right triangles?

Inquiry Activity Summary:

30-60-90

Credits: To the wonderful and beautiful Victoria Ventura

So you may be asking yourself, "what is this hoopla that I am looking at?" Its not as hard as you think or at least how it looks. Let me simplify it for you. The triangle is an equilateral triangle all the sides equal 1. But this means all angles are 60 degrees? So what do I do? You split them in half, making it into two 30-60-90 degree triangles. We know for a fact from our unit circle the bottom "x" side would be 1/2 and the "y" side would be radical3/2. And the hypotenuse would be one.

Now how could this be applied to all different triangles that simply don't have the hypotenuse as one? We'd simply add a variable to make it applicable to all side lengths possible ever! This is the pattern if you haven't caught up...

Below is a picture of an example to help you further on your understanding of 30-60-90 triangles:

Credits: To the wonderful and beautiful Victoria Ventura

45-45-90

Credits: To the wonderful and beautiful Victoria Ventura

Much simpler than the 30-60-90 triangle is the 45-45-90 triangles. Just as we did before with the the equilateral triangle, we would have to split the square in half. Splitting directly through two 90 degree angles. Thus, you'll get two 45-45-90 triangles. Fabulous, I know. All sides are equal to one. But what would the hypotenuse equal? This is where our Pythagorean theorem would take play. 1^2 + 1^2=2 which in fact would be radical2. To make it relate-able to other 45-45-90 triangles, just as we did with 30-60-90 triangles, we would multiply all sides with n.

With this added variable, it would be a great pattern to follow and find other missing sides, knowing limited information.

Here's a fabulous example further showing solving 45-45-90 triangles having only limited information!:

Credits: To the wonderful and beautiful Victoria Ventura

Inquiry Activity Summary:

Something I never noticed before about special right triangles is how simple they can actually be. I made my life misery last year I over-thought things and kind of just got tired so I memorized the sides. I never put thought into how the sides were derived. This I/D really helped me to see that in a clearer light and understand how these triangles actually got the sides.

Being able to derive these patterns myself aids in my learning because I'm learning to be able to solve things on my own without help of anyone else. Of course, just like any other human on planet earth, I get a little help from my friends from time to time but, only when most needed. Just like the previous I/D, it really helps students to learn to actually practice thinking, not just doing careless work and understanding why, what and how and a problem can be solved, such like this I/D. It really helps us see our strengths and weaknesses and is a major help to students.

example: Unit O Concept 6

I/D#1: Unit N Concept 7:The Unit Circle and the Magic Five

Inquiry Activity Summary:

Angles on the first quadrant

30 degree angle

Figuring out the 30 degree angle triangle was very simple, though at first it was a bit difficult. The reason being because I didn't know the measurements of each side. But once I logged onto legoogle.com, I was able to find all the sides. Side R = 2x, side Y = x and side X = xradical3 (as seen in the picture.) Our hypotenuse/sideR had to equal one, therefore, we divided out 2x from the 2x and we got side R to equal 1. Simple right, are you done yet? Nope. In order to divide out the 2x from sideR we had to do that to all the sides. SideX when we divided and simplified it gave us radical3 over two. For sideY we did the same and the x's cancelled therefore leaving us with 1/2.

The second part of the 30 degree angle was finding the points as if it were on a graph. Taking that initiative we drew out the x and y axis. Our first point on the graph was the center point and that was simply just (0,0). The second point on the the x-axis, on the right, would turn out to be (radical3/2, 0) because of the information we had solved earlier. According to rise over run, our last point would have to be (radical3/2, 1/2). Why? Because we went over radical3/2 and rose 1/2. Lastly we filled in the blanks for r, x, and y.

45 degree angle

Figuring out the 45 degree angle was much simpler than the first angle we had to figure out because we already had an idea of how to solve it out. According to our sources, google.com, sideR was xradical2, sideX and sideY were both x. Just as we did for angle 30, sideR had to equal 1. Therefore, we divided out xradical2 from xradical2 to get one. We divided out xradical2 from all the sides. When dividing out xradical2 from x we ended up with 1/radical2, of course we all know, that no radical can be in the denominator, so we multiplied radical2 to both the bottom and top. Our final result was radical2/2. From there we drew the x and y axis and figured out the points, resulting with what is above in the picture. Lastly, we filled in the r, x and y blanks.

60 degree angle

While we all solved this angle we slowly realized that everything was the exact same as the 30 degree angle except everything was reversed. SideR was the same. But the X for the 60 degree angle was 1/2 instead of being radical3/2 and sideY was radical3/2 instead of 1/2. As for the points as well! Instead of going through all the work, since we already knew how to do it from the previous 30 degree angle, we just reversed everything and finished our 60 degree angle in seconds.

How did the activity help us derive the Unit Circle?

Well, glad you asked. These three angles are key to finding the rest of the Unit Circle. All we need to know is each and every angle (referring to 30, 45,  60 degree angle) and we can find the rest of the unit circle.

http://fac-web.spsu.edu/math/edwards/1113/unitcircle.htm

As seen from the Unit Circle above, the only points that are repeatedly seen are 1/2, radical3/2 and radical2/2. That's because all angles have the same common thing, they all have reference angles of 30, 45, and 60. 30 and 150 reflect off one another and have the same point except for the negative x, but that makes sense since it is on the negative side of the x-axis. This method goes to the rest of the angles as well they all reflect one another and tie back to either a 30, 45 or 60 degree angle. The only thing that changes throughout the Unit Circle is the degrees and radians but that's simply because as we go around the circle the angle gets bigger and bigger.

Quadrant II (30 degree angle)

This angle above in fact is the 150 degree angle. But it has the reference angle of 30 degrees. The only thing that really changes compared to the original 30 degree angle is that the 150 degree angle has a negative radical3/2 and of course the degree and radian are different as well.

Quadrant III (45 degree angle)

The angle above is actually 225 degrees. This angle has a reference angle of 45 degrees. The only thing that really changes compared to the original 45 degree angle is that the 225 degree angle's the x and y points are negative. To conclude, the degree and radian are one of the other factors that are different.

Quadrant IV (60 degree angle)

The angle above is actually 300 degrees. This angle has a reference angle of 60 degrees. The only thing that really changes compared to the original 60 degree angle is that the 300 degree angle's x and y points are negative. Lastly, the degree and radian are one of the other factors that are different.

Inquiry Activity Reflection:

The coolest thing I learned from this activity was that I finally understood how to solve the Unit Circle! Last year, in Algebra II, when we were learning this, I was so lost. I wanted to love the Unit Circle but the Unit Circle didn't love me. It was a very depressing time for me. But now me and the Unit Circle are on the same terms. I get it and the circle gets me. It's a happy world isn't it? Yes, very.

RWA#1-Unit M: Ellipses

Ellipses

Definition: The set of all points such that the sum of the distance from two points is a constant. (http://www.lessonpaths.com/learn/i/unit-m-conic-section-applets/ellipse-drawn-from-definition-geogebra-dynamic-worksheet)

http://www.clausentech.com/lchs/dclausen/algebra2/ellipses.htm

Algebraically:

Ellipses can be algebraically seen as the pictures above. An ellipse can be seen in two ways, "fat" and "skinny". How do we determine which is which? Well if the larger denominator, a^2, is under the x it will be a horizontal ellipse or "fat" ellipse. If the larger denominator, a^2, is under the x it will be a vertical ellipse or "skinny" ellipse. "A" give us the indication of the length of the major axis and the "b" gives us the indication of the length of the minor axis. With our "h" and "k" we can find our center. Do keep in mind that the "h" goes with "x" and "k" will always go with "y". With our equation we can find about everything needed to graph our ellipse. To find our a, b and c, we use "a^2-b^2=c^2".

By looking at the equation, we can find the "a" and "b". "A" pertains to our vertices, "b" to our co-vertices and "c" to our foci. Depending whether its horizontal or vertical, our vertices and foci have something in common, either the "x" or "y" for both vertices and foci will be alike. In finding our vertices if the bigger denominator(a^2) is under the y (vertical/"skinny" ellipse), our vertices and foci would have the similar x's. So what would we do with "a"? You add and subtract that to the "y" of the center two get your two vertices. This would be the same process for the "c" and foci. Finding our co-vertices would mean that the "y" would stay the same for both co-vertices. So in this case we would add and subtract "b" to our center to find our two co-vertices. This whole process would be the same for horizontal/"fat" ellipses except vice-versa.

http://blue.utb.edu/qttm/valencia/algebra2/ellipse/ellipse_parts.jpg

Graphically:

Graphically, ellipses look like the picture above. The vertical/"skinny" ellipses consists of the same parts as the horizontal/"fat" ellipses. The center of our ellipse is found in our equation, as mentioned above. With our "h" and "k" we can find our center. Do keep in mind that the "h" goes with "x" and "k" will always go with "y". Our graph consists of a center, vertices, co-vertices, foci, a minor axis, a major axis and eccentricity. To determine the major axis and minor axis would only mean looking at the vertices and co-vertices. Looking at the vertices, whichever, x or y, is similar that would be the major axis. For the co-vertices, whichever, x or y, is similar that would be the minor axis. The only factor that has not been explained is eccentricity. Eccentricity is "c" divided by "a". This determines how much the ellipse deviates. If it's closer to one, it means that it looks more like a circle and vice versa if its closer to zero (0 < e < 1). Eccentricity is not necessarily seen on the graph but it helps the student to get an estimate of how the graph might look like when they draw it.

Here's more help with ellipses:

http://www.youtube.com/watch?v=5nxT6LQhXLM

Real World Applications (RWA)-

http://oneminuteastronomer.com/8626/keplers-laws/

Ellipses can be represented in the orbit of the planets in outer space. How do you say so? Looking at the picture demonstrates it perfectly and why. The sun would always serve as one of the foci. The other focus would be empty but still be there (http://oneminuteastronomer.com/8626/keplers-laws/). As described in the Kepler Law, "their orbits are similar ellipses with the commonbarycenter being one of the foci of each ellipse. The other focus of either ellipse has no known physical significance. Interestingly, the orbit of either body in the reference frame of the other is also an ellipse, with the other body at one focus."(http://en.wikipedia.org/wiki/Ellipse#Ellipses_in_optimization_theory)

The things remaining around the ellipse have no physical "significance" but are still there. The basic real world of application of the ellipse is centered around the foci. One thing that is obvious from the picture above is the foci and the planet. It goes along with the concept of d1+d2= a constant. Which is what ellipses our all about! The picture above perfectly describes why an ellipse can be applied in the real world with the orbit of planets from outer-space.

URL's:

pictures-  http://www.clausentech.com/lchs/dclausen/algebra2/ellipses.htm

               http://blue.utb.edu/qttm/valencia/algebra2/ellipse/ellipse_parts.jpg

               http://oneminuteastronomer.com/8626/keplers-laws/

video-     http://www.youtube.com/watch?v=5nxT6LQhXLM

websites-http://www.lessonpaths.com/learn/i/unit-m-conic-section-applets/ellipse-drawn-from-definition-geogebra-dynamic-worksheet

              http://www.clausentech.com/lchs/dclausen/algebra2/ellipses.htm

              http://oneminuteastronomer.com/8626/keplers-laws/

              http://en.wikipedia.org/wiki/Ellipse#Ellipses_in_optimization_theory