6-3 Ellipses

I am going to show you how to sketch ellipses. There are 7 easy steps on how to get the information so you can sketch your ellipse. I will show you an example problem and step by step how to solve the ellipse and how to sketch it. The steps are really easy to follow and i will also explain each step. The most important part of an ellipse is to know the standard form which is x^2/a^2 + y^2/b^2= 1. If a^2 has the bigger number then x is major axis. if b^2 has the bigger number then y is the major axis. once you find which one is major axis and minor axis then you are ready to finish the 7 steps and continue to graph the ellipse which i will demonstrate in the example problem. The other thing that is really important to do is to graph correctly or else your answer will be completely wrong.


Example: x^2/9 + y^2/16= 1 Sketch the graph.
step 1: find major axis: y
step 2: find minor axis: x (because a^2 is smaller than the b^2)
step 3: take the square root of the largest denomenator: square root of 16= 4. then you put in point form. if major axis is x is goes in x spot and vice versa. (0,4) (0,-4)
step 4: take the square root of the smallest denomenator: square root of 9= 3 (3,0) (-3,0)
step 5: find length of the major axis: 2* square root of largest denom: 2* square root of 16= 2(4)= 8
step 6: find length of the minor axis: 2* square root of smallest denom: 2* square root of 9= 2(3)= 6
step 7: find the focus. the focus goes on the major axis. smallest denom=largest denom - focus^2: 9=16 - f^2= the square root of 7. point form: (0,square root 7) (0,-square root 7)


Now that you have havethe information it's time to graph.

 

6-2 Circles

Dear BRob, this is Danielle Sharp from 1st hour. I never got your e-mail, and i thought that I could use someone else's link but it didn't work, so I'm posting in Sarah's blog cause she's at my house and she let me on it. Anyway, here it goes.

The standard form of a circle is (x-h)^2 + (y-k)^2 =r^2. (h,k) is the center and r is the radius. To find the radius you can use the distance formula with the center and point on the outside or you can half the diameter found by using two points on the outside through the center.
To find the equation of a circle you must put it in standard form by completing the square.
1. Divide by leading coff., then move all numbers to the right
2. Half the middle term then square it and add both sides
3. Factor
When you need to graph in your calculator you need to slove for y and put a positive and negative formula in for y equals.
When you find the intersection of a line and circle
- solve the line for y
- plug circle equation for y
- slove th equadractic
- if you get i they DO NOT INTERSECT

x^2 - 4x + y^2 + 6y + 4 = 0

subtract 4 from both sides (step 1)
x^2 - 4x + y^2 + 6y = -4

(step 2) half the middle term then square and add to both sides.
-4 dvided by 2 equals (-2) then square that and it equals 4
6 divided by 2 equals 3 then square that adn it equals 9
x^2 - 4x + 4 + y^2 + 6y + 9 = -4 + 4 + 9

(step 3) factor
(x-2)^2 + (y+3)^2 = 9
*If you have trouble factoring the equation will always lead back to the number after dividing but before squaring

Your center is (2,-3) and your radius is 3

And that is how you slove for a circle, hope you enjoyed :)
-- Danielle Sharp!

6-3 ELLIPSES

Ch.6, Section 3-Ellipses
Before you can begin sketching an ellipse, you must either complete the square or divide to get your equation in standard form.


The first step in sketching an ellipse is finding your major axis. Your major axis will be the variable with the largest denominator.

Next, you will state your minor axis. Your minor axis is the variable with the smallest denominator.

Now you will find your vertex. You start by finding the square root of the largest denominator. After you've done that, you must put it in point form. If the major axis is x, then your answer will go in the x coordinate spot. If your answer is y, then it goes in the y coordinate spot.

To find your other intercept, you must square root your smallest denominator. Your answer will go in point form just as your vertex.

Next, you have to find the length of your major axis. To do this, you must multiply 2 by the square root of the largest denominator.

Now you find the length of your minor axis. You do the same as you did to find your major, except you multiply 2 by the square root of your smallest denominator.

The last thing you do before actually sketching your ellipse is finding the focus. To find the focus, you use the equation, smallest denominator = largest denominator - focus^2(squared). Your focus will always be a point on the major axis.




EX: (x^2/4)+(y^2/9)=1

1) y is major because is has the larger denominator

2) x is minor because it has the smaller denominator

3) square root of 9 is +/-3, vertex =(0,3) (0,-3)

4) square root of 4 is +/-2, other intercept =(2,0) (-2,0)

5) 2 x square root of 9 = 6, lenth = 6

6) 2 x square root 4 = 4, length = 4

7) 4 = 9 - focus^2
= square root of -5 = square root of focus^2
= +/-2.2
Focus = (2.2,0) (-2.2,0)


And this is how to solve an ellipse :) -JORDAN

6-4 Hyperbolas

Standard Form of Hyperbolas:

x^2/a^2 -y^2/b^2 =1 or -x^2/a^2+y^2/b^2 =1

The steps for solving a hyperbola are as follows:

1. Identify the major axis which is always the largest denominator.

2. Find the minor axis which is the smallest denominator.

3. Take the square root of the largest denominator and write it in point form (x,y) and (-x,-y)

4. Take the square root of the smallest denominator and write that in point form.

5. To find the focus, one must use the formula f^2 equals largest denominator plus smallest denominator. Write the focus in point form when it's found.

6. To find asymptotes one must use the formula y equals plus or minus the square root of the y denominator over the square root of the x denominator. It is to be written in slope intercept form.

An example of this problem

x^2/81 - y^2/9=1

1. The major axis is x.

2. The minor axis is y.

3. Take the square root of 81 and one will end up with 9. In point form it looks like (9,0) and (-9,0)

4. Take the square root of 9 and one will end up with 3. In point form, it's (0,3) and (0,-3)

5. The focus formula in hyperbolas is focus squared equals largest denominator plus smallest denominator, the f^2= 9 + 81 which makes focus squared equal 90. When you square root both sides you get focus equals the square root of 90 which simplifies to 3 square roots of 10. It goes in point form simplified.

6. The square root of the y denominator (9) is put over the square root of the x denominator(81). The asymptote is placed in slope intercept form as y= +or- 1/9x

And that's the way the cookie crumbles!

-Sameer

6-3 Ellipses

6-3 Ellipses
This week I am going to explain how to do ellipses. There are seven steps that are required to solve an ellipse.

The standard form of an ellipse is:

You can do one of two things to put an ellipse in standard form. The first is to divide and the second is to complete the square.
Steps:
1. The first thing that you have to do is to state the major axis. You can find the major axis by looking for the variable with the largest denominator.
2. Next you have to state the minor axis. You can find the minor axis by looking for the variable with the smallest denominator.
3. Then you must find the vertices. You can do this by taking the square root of the largest denominator. This gives you a positive and negative answer. You then put that into point form. If the major is x then the numbers you get go into the x coordinate. If the major is y then the numbers you get go into the y coordinate. You should now have two points.(x,0)(-x,0) or (0,y)(0,-y)
4. Now you have to find the other intercepts. To do this you take the square root of the smallest denominator. Again, you get a positive and negative answer to put in point form. You put these numbers in the coordinate of the minor axis. (x,0)(-x,0) or (0,y)(0,-y). You should now have four points total.
5. To find the length of the major axis you multiply 2 * the square root of the largest denominator. This will give you one number.
6. To find the length of the minor axis you multiply 2* the square root of the smallest denominator. This will also give you one number.
7. Time to find the last piece of the puzzle, the focus. To find it, follow this formula: smallest denominator=largest denominator – focus^2. It will give you a positive and negative number to put in point form. The numbers go in the coordinate of the major axis. (x,0)(-x,0) or (0,y)(0,-y).
After you finish all of the steps you have to sketch the graph.

Example
See numbered steps above for details of each step.
(x^2/4)+(y^2/9)=1
1. Y is the major axis
2. X is the minor axis
3. Square root of 9= +/- 3 (0,3) (0,-3)
4. Square root of 4=+/- 2 (2,0)(-2,0)
5. 2 * square root of 9= 2*3= 6(length of major axis)
6. 2 * square root of 4= 2* 2= 4(length of minor axis)
7. Focus: 4=9-f^2
-f^2=-5
f^2=5
f= +/- 5^1/2 (0, 5^1/2) (0,-5^1/2)
Here is the sketch of the graph: (The red points are the vertices and the green points are the foci)

Intersection of circlesssss!

So today we will be learning about how to find the intersection of a line and a circle when given a line and cirlce.

1. the first thing your going to do is take the line equation and solve for y.

2. take the line equation, solved for y, and plug it into the cirlce equation for y.

3. The next step is to combine all the terms and distibute the powers and then use foil to help expand .

4.Then you solve the quadratic and will get 2 x's. X will equal one number, and then X will equal another number.

5. Next you will take one of those numbers and plug it into the line equation as an x and solve and that will be one y. Take the other number and do the same and that will be your second y.

6. you will have 2 points and that will be where the line intersects the circle. If you get an i then that mean it doesnt not intersect.


Ex: x+y=23, x^2+y^2=289

1.y=x-23

2. x^2+(x-23)^2=289

3. x^2+x^2-46x+240

4. x=8 x=15

5. 23-8=y y=15 23-15=y y=8

6. (8,15) (15,8)

The line will intersect the circle at (8,15) and at (15,8).

6-3 Ellipses

I am going to explain how to do ellipses. There are seven steps to working an ellipse.

Standard form of an ellipse is as follows:

Put an ellipse into standard form by dividing or completing the square.

1. 1. .Define the major axis (The variable with the largest denominator is major)

2. 2Define the minor axis (The variable with the smaller denominator is minor)

3. 3. Find the vertex. The vertex is the square root of the major axis (larger denominator). It is put into point form. The answer goes into the slot of the major axis.

4. 4.Find the other intercept. The other intercept is the square root of the minor axis (smaller denominator). It is put into point form and the answer goes into the slot of the minor axis.

5. 5 Find the length of the major axis. This is found by multiplying the square root of the major axis by two.

6. 6.. Find the length of the minor axis. This is found by multiplying the square root of the major axis by two.

7. 7. Find the focus. (The focus is ALWAYS on the major axis.) It is in point form. The equation for finding the focus, vertex, or other intercept is smallest denom=largest denom-focus^2.

EXAMPLE

Sketch the ellipse.

X^2/25+y^2/169=1

1. . Y is major.

Again, the larger denominator is the major axis.

2. 2. X is minor.

The smaller denominator is the minor axis.

3. 3. Vertex=square root of 169=+/-13 (0,13) (0,-13)

The vertex will always be the square root of the major axis put into point form.

4. 4. Other intercept=square root of 25=+/-5 (5,0) (-5,0)

The other intercept is always the square root of the minor axis put into point form.

5. 5. 2*square root of 169=26

The length of the major axis (in this case y) is found by multiplying 2*the square root of the larger denom (in this case 169) which would make it 2*13=26.

6. 6. 2*square root of 25=10

The length of the minor axis (in this case x) is found by multiplying 2*the square root of the smaller denom (in this case 25) which would make it 2*5=10.

7. 7. 25=169-f^2

f^2=144

f=+/-12 (0,12) (0,-12)

The focus is found using the equation smaller denom=larger denom-f^2. You first subtract 169 from both sides. Then you square root both sides to get +/-12 and put it into point form. THE FOCUS IS ALWAYS ON THE MAJOR AXIS.

The graph would look like this.

That’s an ellipse.

--Sarah

Parabolas!

I am going to be showing how to do something we’ve learned in the past week. It can be hard at first but after you get the hang of it, it'll be really easy. We’re going to be working on solving Parabolas. There are 5 simple steps to working these types of problems. The only complicated part is that you do everything opposite as what you would think.

The first thing I’m going to show you is the 5 steps:

Step 1: The first step is going to be to find out what direction the Parabola is going to be facing. If you have an x^2 it will be facing up, if you have a –x^2 it will be facing down, if you have a y^2, it will be facing towards the positive side of the x-axis, and if you have –y^2, it will be facing towards the negative side of the x-axis

Step 2: The second step is to find the axis of symmetry. You do this by using the formula x=-b/2a. This will also be used to find the vertex.

Step 3: The third step is to find the Vertex. You do this by using the formula ((-b/2a, F(-b/2a)). Here’s were it can get confusing. If you have a x^2 term you will find the x coordinate, then you will plug that number into the equation. This is the equation shown above. If you have a y^2 term, you would do the same thing except you would switch the x and y coordinates.

Step 4: The fourth step is to find the focus. You do this by using the formula (1/4p)=leading coefficient. Once you find p, you use the formula, Vertex coordinate+P=Focus. This is where it can get tricky. Your vertex coordinate is probably opposite of what you are thinking. If you have an x^2 term, you will use the y coordinate and if you have a y^2 term, you will use the x coordinate. (this is written in point form)(Something that helps me remember which one to use is if there is an x^2 term, the only thing you do with the x’s is carry them down through the steps and use the y’s for everything else. If you have a y^2 term, the only thing you do is carry the y coordinates down through the steps and everything else is done using the x’s)

Step 5: The last step is going to be to find the Directrix. Since you already found p in the previous step, the is very simple. All you do is take your Vertex Coordinate and subtract p using the formula, vertex coordinate-p=directrix. (make sure this is in equation form).

I will work an example for you using the same steps as written above.

Example: Solve the following Parabola, y=(1/6)x^2

Step 1: This will be a parabola opening upwards.

Step 2: Using the equation x=-b/2a, x=-0/2(1/6) This gives you x=0, therefore, your axis of symmetry is x=0

Step 3: Because you have a x^2 term you will use ((-b/2a, F(-b/2a)). Your x-coordinate is going to be the same as your axis of symmetry. So you know that your Vertex is now (0, F(-b/2a)). Now you will plug 0 into your original equation and get y=(1/6)(0)^2 and get y=0. You now know that your vertex is (0,0)

Step 4: You automatically know that your x term is going to be 0 because it never changes when going through the steps. Now you have 1/4p=leading coeff. 1/4p=1/6, so p=(3/2) now you use the formula vertex coordinate+p= focus, so 0+(3/2)=3/2. This means that your focus is (0,(3/2))

Step 5: Since you already have P, you just use the formula Vertex Coordinate-p=directrix, so 0-(3/2)=(-3/2) This means that your focus is y=-3/2

I hope you learned something!

--Carley(:

6-4 Hyperbolas

The standard form of a hyperbola:

The Steps for solving a hyperbola:

• The larger denominator is the major axis.
• The smaller denominator is the minor axis.
• Vertex = the square root of the largest denomiator
• Other value = the square root of the smallest denomiator
• Find the focus with the equation f to the second = larger denominator + smaller denominatior. Then f to the second = vertex to the second + the other value to the second.
• Find the asymptotes. Equation: y= +- the square root of the y denominator over the square root of the x denominator.

Example:
Solve the hyperbola.

9x² - 16y² = 144

First, you divide everything by 144. Which will leave you with:
x²/16 - y²/9 = 1

After you divided everything by 144, you officially start the steps from above. You need to state what is your major and minor axis.The major axis has to be positive. One will be negative (either x or y, it doesn't matter).
Major axis: x
Minor axis: y

After you have found the major and minor axis, you need to take the square root of your major axis.
Which would be the square root of 16 = +-4. Your points will be (4,0) (-4,0).

Then you do the same thing for your minor axis. Square root of 9 = +-3. Points: (0,3) (0,-3).

Now you have to find your focus. To find your focus, you use the equation focus squared = larger denominator + smaller denominator. F^2 = 16 + 9 = 25. You take the square root of your answer which is 25. So the square root of 25 = +-5. Focus: (5,0) (-5,0).

Your last step is to find the asymptotes. You use the equation in the last step from the steps above that I listed. Y = square root of 9/square root of 16 = 3/4x. Your finally answer for your asymptote is Y = +-3/4x.

Amber Smith

Okay so, I'm going to explain how to do parabolas.

There are 5 easy steps in working out parabolas. These steps are:

1. The first thing you would do is look at the problem. Lets use two different problems as an example, one with y= and one with x=. This way you can understand it both ways. The first example is going to be (example 1)y=2x^2 and the other example is going to be (example 2)x=-2y^2. The first thing your going to do is look if there is an x or y being squared. In the first example you have a x being squared, where as in the second example you have a y being squared. Since in the first example you have a positive x being squared, you are know going to know that the parabola is going to go upwards. If you were to have a negative x being squared then you would know the parabola would be going down. Now, in the second example you have a negative y being squared, this lets you know that the parabola is going to go to the left. If you were to have a positive y then your parabola would face the right.
2. The second step would be to find the axis of symmetry. The equation to find this is x=-b/2a or y=-b/2a. This equation is going to depend on the problem you are working. If your probelm is like example 1 then you are going to use the x= equation and vice versa. So for example, if you were working example 1 and got x=h, the h would now go in point form (h, ?) The x-coordinate is never going to change throughout the whole problem. Now if you were working example 2 and got y=d, you would have (?, d) and that means the y-coordinate is never going to change throughout the whole problem.
3. The next thing you want to do is to try and find the missing coordinate. This step is very easy. All you do is plug in what you got for the step above into the equation. So for example, if you were on example 1, where you got x=h, you would then plug h into the equation which would be y=2(h)^2. Whatever you would get for that would be your other coordinate. So lets just say you got w. Your vertex would then be (h, w). The same goes for the other example, you would just switch up the x and y.
4. This next step is to find the focus. To find the focus you need to use the equation 1/4p=coefficient of leading term. So lets use example one, the leading term would be 2. So once you work that out, lets just say you got p=R So once you would get your answer, you would then add to your other vertex coordinate. REMEMBER: the one that you solved for in step 2 NEVER changes. So you would do (w+R) and whatever that would equal would be your focus. Vice versa goes for example 2.
5. The last step is going to find the directrix. This step is really easy. You would do the other vertex coordinate - p and that is going to equal the directrix. You would then write whatever you got for that as y=the # or vice versa for a x= problem.

Note: The vertex is going to be right in the middle of the focus and directrix.

Now, I'm going to work out a full example for you. I'm going to number the steps in my work to match the ones above.

EXAMPLE: y=-2x^2

1. The parabola is going to go downwards.
2. x=-0/2(-2), which is going to be x=0.
3. y=-2(0)^2, which is going to be y=0. You are then going to have a vertex of (0,0).
4. 1/4p=-2. Your going to multiply 1/4p to both sides, getting -8p=1, which then gives you p=-1/8. You would then do 0+(-1/8)=-1/8. The 0 you add is going to be the one that doesn't change the whole problem which is the y-coordinate. Your focus is then going to be (0,-1/8).
5. 0-(-1/8)=1/8. Which would then give you y=1/8. Which is going to be your directrix.

This is what the graph is going to look like:

And that's how you work parabolas.

Halie (:

Welcome!

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