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Contents

- 1 X-intercepts
- 2 X-intercepts
- 3 Y-intercepts
- 4 Y-intercepts
- 5 Ellipses at the Origin
- 6 Graphing Ellipses Check 1
- 7 Graphing Ellipses Check 1
- 8 Graphing Ellipses Check 2
- 9 Graphing Ellipses Check 2
- 10 Moving an Ellipse
- 11 Center of an Ellipse
- 12 Center of an Ellipse
- 13 General Equation for an Ellipse
- 14 Ellipse Practice 1
- 15 Ellipse Practice 1
- 16 Ellipse Practice 2
- 17 Ellipse Practice 2
- 18 Ellipse Practice 3
- 19 Ellipse Practice 3

We've looked closely at circles, so now let's examine ellipses. Ellipses are very much like circles. There's just one exception. The distance we go in the x direction will be different from the distance we go in the y direction. So, sometimes we might move further from the center in the x direction than we would vertically in the y direction. And other times we might go further in the y direction from the center, than we would in the x direction from the center. We'll see a lot of similarities between circles and ellipses, so lets try looking at a circle first. There's this amazing online graphing calculator called Desmos. If you've never heard of it, I highly recommend that you check it out. It allows us to easily enter in equations. And we can instantly see what the graph looks like. For now we're going to look at the equation of a circle, but if you ever want to play around with this on your own, go to www.decimals.com. You can create your own account and then start to draw and save your own graphs. For circles we know they have equal amounts of x squared and y squared. And if we take the square root of this number, we'll get six. That's the radius of our circle. The distance to any point from the center to another point on the circle. But what if we wanted to graph this equation? Notice that we have different amounts of x squared and y squared now. Now, I'm going to go ahead and let you know that this will be the graph of an ellipse. So, let's see how we can graph it ourselves. First we'll take the equation and we'll divide through by the number 36. When we divide each term by 36, we'll get one fourth x squared here. one ninth y squared here and 1 on the other side of the equal sign. So, we're going to ignore our first equation of the circle for now. We've taken this equation and rearranged it into this. Now, can you tell me what are the two x-intercepts?

The x intercepts would be 2,0 and negative 2, 0. Great solving if you found those two. Remember for an x intercept, we're somewhere along our x axis, which means the y value must equal 0. So we plug in zero for the value of y, and then we can continue solving for x. Here we have x squared divided by 4 equals 1. So we're multiplying both sides by 4 to clear the fraction and then we'll have x squared equals 4. We'll take the square root of both sides, indicating plus or minus, so x can equal positive 2 or negative 2. Keep in mind that x intercepts are points, so we need to write the coordinates of the x intercepts. We'll have positive 2,0 and negative 2,0. Both the coordinates written in parentheses.

So here's what we know so far. We took the equation of our ellipse and we divided it through by 36. We came up with this equation and we just found that the x intercepts were (2,0) and negative 2,0. So, we know these two points are on our ellipse. So, now that we found the x intercepts, I want you to see if you can find the y intercepts. What do you think those two would be?

The y intercepts would be 0 comma 3 and 0 comma negative 3. Great work if you got those two correct. We know for y intercepts we need to let x equal 0. So we set the value for x equal to 0, and then we solve for y. 0 squared is simply 0, so this fraction with the 0 in the numerator simplifies to 0. Then we'll just have the rest of our equation. Y squared divided 9 equals 1. We'll multiply both sides of the equation by 9 to get y squared equals 9. And then we'll take the square root of both sides using plus or minus to get 1 y is equal to plus or minus 3. Notice that we had a y squared term. This led us to have two solutions for y. We'll have positive 3 and negative 3. We use positive 3 for one of the y-coordinates, and negative 3 for the other y-coordinate. The x-coordinates are both 0.

So, these two points where x intercepts at 2, 0 and negative 2, 0. And these two points are the y intercepts. They're at 0, 3 and 0, negative 3. Now we can connect these points with a smooth curve and then we'll have our ellipse. So, for our circle, we had equal amounts of x squared and y squared and our radius was 6. So, we started from the center and we moved 6 units left and right and up and down to graph that circle. Now, for our ellipse, we're going to move two units left and right in the x direction. And three units up and down in the Y direction. These two points are the X intercepts, and these two points are our Y intercepts. Now, there's an easier way to graph this ellipse than finding the X intercepts and Y intercepts. Remember that this equation is the same as this equation, so let's graph this one instead. I'm going to get rid of this equation first and now I'm going to write this equivalent equation in its place. We'll have x squared divided by 4 plus y squared divided by 9 equals 1. And remember that we're basing this equation on the Pythagorean Theorem. So we can really think of 4 as 2 squared and we can really think of 9 as 3 squared. We could have graphed this ellipse by looking at these two numbers. The squared number underneath the X squared term tells us to move two units left and right from the center. So, since this is a two squared, we move two units to the right from the center, and two units left from the center. The squared number under the Y squared term tells us to move three units up and three units down from the center. This gives us a same four points and we are able to graph our ellipse. In general this is the equation we use for an ellipse, centered at the origin. Notice that there is no number added or subtracted to the x or to the y. So, the center would be at 0, 0. Now what I have gone ahead and done is added this equation into decimals/g. And I've created sliders, or adjustments, for the values of a and the values of b. Notice that if a and b are both 1, we get a circle centered at the origin. This is the idea that we'll have equal amounts of x squared and y squared, and the radius would just be 1. Now, if I increase a, we stretch in the x direction 1 unit to the left and 1 unit to the right. This gives me an ellipse since I have more X squared than I do Y squared. And, we could keep stretching our ellipse. We could have an A value of squared term, that's the direction it stretches in. In order to stretch the ellipse vertically up and down, we want to adjust the b value. We can make the b value be 2, or 3, or even 4. And notice that if we bring it up all the way up to 5, again we'll have a perfect circle. Since we have even amounts of X squared and Y squared. This is why it's helpful to identify the A value and the B value to graph an ellipse. The A value tells us how far we should move from the center left and right, and the B value tells us how far we should move from the center up and down.

Now that you've seen the general form for an ellipse, what do you think the center and the values for a and b, would be for this equation? Put the center here, put the value of a here and put the value of b here.

This ellipse would be centered at the origin and it would have an a value of 6 and it would have a b value of 4. Great thinking if you got each of these correct. We know the center is 0,0 since this fits the general formula for our ellipse. There's no number next to the x inside parenthesis and there's no number next to the y inside our parenthesis. So we know our ellipse is centered at 0, 0. To find the a value we want to take the square root of this denominator of 36. And the square root of 36 equals 6 which means we go 6 units in the left direction and in the right direction. Remember the a term indicated which way we move in the x direction. Since the a number is directly underneath the x squared term. For b, we just need to take the square root of 16, which equals 4. So, to get two other points on our ellipse, we move 4 units up from the center, and 4 units down from the center. This gives us four easily identifiable points, and we can connect our ellipse. Remember that this was our general formula for an ellipse centered at the origin. So we need to find the square root of 36 and the square root of 16. I think it's always more helpful to write these as perfect squares if you can. Comparing the equations like this, we can quickly see that the value of a equals 6 and the value of b equals

So, here's a different ellipse. What do you think would be the center for this ellipse? What do you think the a value is and what do you think the b value is? Keep in mind that for an ellipse, the easiest way to find these values is to set this equation equal to one.

For this equation, the center is at O,O the a value is 10 since we moved 10 units in the x direction, left and right. And the b value is 5 since we moved vertically from the center 5 units up and 5 units down. Now if you're stumped about what to do first, remember that we want this to equal one, so we should divide our equation through by 400. So, we'll take our equation and divide each term by 400 4 divides into 4, leaving us with 1X squared in the numerator. And divides into 16 one time, and 16 divides into 400 25 times. So, we'll have 1Y squared, divided by 25. And finally, on the other side of the equals sign, we'll have one, since 400 divided by 400 equals one. Now we're looking at an equation that fits our general form when the ellipse is centered at the origin. The center is 0, 0, since there's nothing next to the x and nothing next to the y. We simply just have x squared and y squared. Now, the a value is the square root of this number. So, the a value must be 10. And the b value is the square root of this number. So b must be positive 5. If it helps you, you can rewrite the denominators as perfect squares. This easily lets you find the a value and the b value. So, we took this equation, and we changed it to look like something else. We've changed the equation to look like this. Let's graph this one to make sure they're identical. So, so far we have our old blue ellipse just like from before that represents this first equation. And you can see that I'm on it since it's become bold, since I've clicked on it. Now if I take our equation and I set it equal to 1, we have the same exact ellipse. We moved 10 units in the x direction, left and right, and 5 units in the vertical direction, up and down. And we can even write the equation in this form with exact same elipse. These three equations are written in different forms, but they're all equivalent.

Let's take another look at the original ellipse we first encountered at the start of the lesson. We knew that this ellipse was centered at the origin 0, 0, in that we moved two units left and right to get a point and three units up and down from the center to get two more points. But, not every ellipse is going to be centered at the origin. Sometimes they can be elsewhere. Just like we could shift circles, we could also shift ellipses. So, if I wanted to graph this one, I would notice that the graph would shift up one whole unit. Remember, when we replace y with y minus a number, the graph really shifts up that many units. And instead of just shifting up one unit, I could also shift it up two units. We could shift it up three units, four units or even five units. The key thing to remember is that this sign is the opposite of the direction the ellipse moves. So now let's say if we are looking at our original lips and we want to move it down instead of subtracting from y we'll add to y. So if we add 1 to y, we'll get a shift downward. Our entire orange ellipse moved down one unit. And we get this new blue ellipse. And I could continue shifting it further down by adding a larger number, like two, or three, or four and so on. So we know adding or subtracting the y makes a graph, or an ellipse move up or down. So to move a graph left or right, we'll have to add or subtract to x. So again, notice that I have my original graph, and this time I'm going to subtract 1 from x. If I subtract one from x, I'll get this graph. Notice that all the points on our ellipse shifted one unit to the right. Here's just a few of the points that you can see that all shifted one unit to the right. And if we wanted to, we could shift two units to the right, three units to the right, or even four. And again, notice that this is a subtraction sign. So if it's a negative here, we're really moving to the right. But if it's a positive here, we're really moving to the left. So this new equation would be our original orange ellipse, just shifted one unit to the left. And notice that we have plus the left. And plus 3 would shift it 3 units to the left.

Based on what you think about moving an ellipse, what do you think would be the center for this ellipse? When you think you know the center, enter it here.

This ellipse would be centered at 2, negative 1. Good thinking if you found this point. This would be the basic ellipse centered at the origin. So, first we want to subtract 2 from x. And remember, subtracting 2 moves the ellipse to the right 2 units. Now, we just want to add 1 to y. And when we add 1 to y, our ellipse really shifts down one unit. So, again, here was our original ellipse, centered at the origin. That would be this equation. And then, we shifted it two units to the right, since we subtracted 2, and then we shifted it one unit down, since we added 1 to y. Remember that these numbers are the opposite in the direction. So, you might think negative 2 would shift our ellipse two units to the left. But in fact it shifts our ellipse two units to right. Likewise with y, we might think our ellipse would move up one unit, but it turns out our ellipse moves down one unit. This is the same idea that we saw for circles, and it goes against our intuition. You want to be careful about it.

So this is the general formula for any ellipse. The center doesn't have to be at the origin and the values of h and k determine what our center is. Notice too, that if our center is the origin, then the value of h would be 0 and the value of k would be 0. If we subtract 0 from x, we'll still have x. So we'll just have x squared over a squared plus y squared over b squared equal to 1. This is the first general formula we saw for any ellipse centered at the origin. I mean, this was the second general formula we saw for any ellipse centered at h comma k. Keep in mind that if these signs are negative then both the values of h and k will be positive. Whereas these signs are positive then that means the values of h and k are negative.

So, let's see if you can put that general formula to use. What do you think would be the center for this ellipse, and what do you think would be the values of a and b? Enter those three values here.

This ellipse would be centered at 2, negative 1. Nice work if you got the center. The a value was 3, and the b value was 6. Even better if you got all three correct. We know that the center is at 2, negative 1, since this is a minus 2 and this is a positive 1. Since we subtract two from x, we really move two units to the right from the origin. Then we move one unit down since there's a plus 1 with the y term. Remember we move in the opposite direction of what we think. So, now that we know the center is here, we can use the values of a and b to get four other points. To get the value of a, we take the square root of 9, which equals 3. And to get the value of b, we take the square root of 36 which equals 6. We know that we need to move left 3 units and right 3 units from the center to get two other points on our ellipse. We can also move six units up and six units down, to get two more points on the ellipse. Now if you haven't quite got it, that's okay, try this next one.

What do you think for this ellipse? Where's the center? And what's the value of a and the value of b?

This ellipse would be centered at negative 2 negative 3 would have an a value of 4 and a, b value of 2. Great work if you found all 3. For the center we take the opposite signs of what's next the x and the y in the parenthesis. Ellipse so for x we'll have negative 2 for the x coordinate at the center and negative centered at the origin two units left and then three units down. So, this is how we know our center is here. Next we take the square root of 16, which is our a value. Remember a squared is always underneath the x squared term for ellipses. And to find b, we take the square root of the denominator of the y term. So, the square root of 4 equals 2. From our center we can move four units left and right to get these two points on our ellipse. And from our center we can move two units up and down to get two additional points on our ellipse. And if we look closely at the points that are vertically aligned with the center, we can see that one point is two units above the center and another point is two units below the center.

What about this equation? Where do you think this center would be, and what are the values for a and b? Now keep in mind that this equation is not set equal to

This ellipse would be centered at negative 1 comma 3. It'll have an a value of can find this center by looking at the x squared term and the y squared term. The x is a plus 1 x to it so that means the x coordinate of the center will be negative 1 and the value y has a negative 3 or subtraction of 3 next to it. So, this means that the y coordinate of the center is positive 3. Always keep in mind that the values of h and k for our center are the opposite of these two signs. This will always be true for circles and ellipses. So, now we need to find the values of a and b. So, what we should do is take our equation and divide through by 100. That will set our equation equal to 1. So, here's my original equation and now we're going to divide each term by 100. A 100 divided by a 100 equals 1, so this will simplify. An then we have 4 divided by a 100. Well 4 divided by 4 equals 1. An 4 divided it into 100, equals 25. So, we'll have 1 in our numerator, with the y minus 3 squared divided by 25 and the denominator. I don't write the 1 here because it's just multiplication, 1 times any expression is itself. And then finally, on the right hand side, we'll just have 1. Now we want this equation to look like this one. But notice I don't have an a squared term, I don't exactly have a fraction here but really I do. Remember we can divide any expression by 1, dividing by 1 will give us this same exact expression. So, now I can rewrite these denominators with 1 squared equally 1 and 5 squared equaling 25. This is how we get the values away and b, and thinking back to our center. The value of h really needs to be negative 1 in order for the equation to have positive 1. And the value of k would need to be 3 since we'll have 3 here. So, we can see that our center is at negative 1, move 5 units up and down to get two more points on the ellipse. Highlighting these points, we can see that these two are 5 units above and below our center, and that this point is 1 unit to the right of our center. Now desmos isn't showing this point, but that's just because it only shows us certain points of interest. Now, there a reason why it's only showing us these points of interest. And when you learn more about ellipses, you'll discover why these points are so important. For this one, it happens to be a y intercept. And, in fact, desmos will also tell us the x intercepts. These other two points are really important and perhaps you can do some research to figure out why.