Hyperbolas

This week in Adv. Math 1^st hour we learned all about conics. There 4 types of conics: Circles, Parabolas, Hyperbolas, and Ellipses. The conic that I am going to talk about is a hyperbola. Hyperbolas and Ellipses have the same standard form but are different in the way that hyperbolas include a negative sign in its form and an ellipse does not. The standard form for a hyperbola is:

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

Steps :

1. The major axis/largest deno. is the variable that is positive.

2. The minor axis/smallest deno. is the variable that is negative.

3. The vertex is the square root of the largest deno. (put in point form.)

4. The other value is the square root of the smallest deno. (ignore the negative; put in pt. form.)

5. To find the focus, use one of these 2 formulas:

F^2= largest deno. + smallest deno. or F^2= v^2 + other value^2

6. To find horizontal asymptotes use: y= square root of y deno. / square root of x deno. * X

Example Problem:

-X^2 / 144 + Y^2 / 16 = 1

1. Y is major axis ( it is positive)

2. X is minor (it is negative)

3. Square root of 16 (major axis) = +/- 4 (0,4) (0,-4)

4. Square root of 144 (minor axis) = +/- 12 (12,0) (-12,0)

5. F^2= largest deno. + smallest deno.

F^2 = 16 + 144= 160. f^2=160. Square root both, f= square root of 160. (0, Square root of 160)

6. Y= square root of 16 / square root of 144. = 4/12 = 1/3. Y= +/- 1/3x.

Hyperbola

This week we learned about hyperbola's as part of chapter six. The standard form for a hyperbola could be x^2/a^2-y^2/b^2=1 or -x^2/a^2+y^2/b^2=1.

1. The largest denominator is the variable that is positive.

2. The smallest denominator is the variable that is negative.

3. To find the vertex you square root the largest denominator then put in point form.

4. Square root the smallest denominator to find the other value but pay no attention to the negative.

5.&6. We do not do these steps for this section.

7. To find the focus or denominator use the formula: focus^2=largest den.+smallest den. Or focus^2=vertex^2+other value^2

8. To find the horizontal asymptote use the formula y=+/- square root of y den. Over the square root of x den. X

Example. X^2/25-y^2/4=1

1. X is major
2. Y is minor
3. Square root of 25 is +/-5 (5,0) (-5,0)
4. Square root of 4 is +/-2 (0,2) (0,-2)
7. Focus^2=25+4
F=+/- square root of 29. (square root of 29,0) (-square root of 29,0)
8. Horizontal asymptote: y=square root of 4/square root of 25 x
y= +/-2/5x

Ellipses!

This week in math we learned about Conic Sections. One conic section is called an Ellipse, and i will show you an example of how to solve an equation of an ellipse.





(x^2/a^2)+(y^2/b^2)=1



1.find the major axis
2.find the minor axis
3.find the vertex
4.find the other intercept
5.find the length of the major axis
6.find the length of the minor axis
7.find the focus


(x^2/64)+(y^2/144)=1


1. the major axis is y.
2. the minor axis is x.
3. sqrt(144)=12 (0,12) (0,-12)
4. sqrt(64)=8 (8,0) (-8,0)
5. 2* sqrt(144)=24
6. 2*sqrt(64)=16
7. 144=64+f^2; f^2=80; f=sqrt(80) (0,sqrt(80)) (0,-sqrt(80))

Ellipses :D

In class this week we learned about conic sections. I will tell you about Ellipses. The standard form for an Ellipse is x^2/a^2 + y^2/b^2 = 1. To put in standard form you either divide or complete the square.

1. The major axis is the variable with the largest deno.

2. The minor axis is the variable with the smallest deno.

3. The vertex is the square root of the largest deno. To put in pt form, if the major axis is x the answer goes in the x spot and vice versa.

4. The other intercept is the square root of the smallest deno. Then put in pt form.

5. The length of the major axis is 2 X the square root of largest deno.

6. The length of the minor axis is 2 X the square root of the smallest deno. The focus is always on the major axis and is a pt.

7. To find the focus, vertex, or the other intercept use the formulas below. Smallest deno = largest deno – focus^2.OR Other intercept^2 = vertex^2 – focus^2.

Ex.

1. x^2/25 + y^2/4 = 1

1. x

2. y

3. +/- 5 (5,0) (-5,0)

4. +/- 2 (0,2) (0,-2)

5. 2 X square root of 25 = 10

6. 2 X square root of 4 = 4

7. 4 = 25 – f^2

-25 -25

-21 = -f^2

f^2 = 21

f = square root of 21

(square root of 21,0) (-square root of 21,0)

Circles :3

Wazzzam yew guiseeee ;). I bet y'all all can't wait for school Monday, YAAAAYY.

Give an equation of a circle with end points on a diameter (2,1) (12,1).

We know that the standard form of a circle is: (x-h) ^2 + (y-k) ^2=r ^2.

Figuring out the distance from one end of the circle to the other, we will need the distance formula. (to figure out the diameter which will result in finding the radius).

D: √(x2-x1) ^2 + (y2-y1) ^2. You plug in the end points of your diameter (which was given) into this formula:

D: √(12-2) ^2 + (1-1) ^2 = √100 =10

D=10, R=5. So now we know that: (x-h) ^2 + (y-k) ^2 = 5 ^2

H and K in your standard formula is the center coordinates. To get this, you use the midpoint formula (hence center/midpoint).

Midpoint formula= (x1+x2 /2 , y1+y2 /2)

Plug in coordinates from original end points of diameter and get:

(2+12 /2 , 1+1 /2) = (7,1). When plugging this into your formula, remember that the coordinate’s signs change as shows:

Answer: (x-7) ^2 + (y-1) ^2 = 25 (25 comes from the radius which was found from using the distance formula)

Kthatisall.

Standard form of an ellipse…. X^2/a^2 + y^2/b^2 = 1.

1. The major axis is the variable with the largest denominator under it.

2. The minor axis is the variable with the smallest denominator under it.

3. The vertex is sqrt largest denominator.

4. The other intercept is sqrt smallest denominator. Put 3 and 4 in point form.

5. The length of the major axis is 2 times sqrt largest denominator.

6. The length of the minor axis is 2 times sqrt smallest denominator.

7. To find the focus, vertex, or other integer: smallest denominator = largest denominator - focus^2.

Example: x^2/100 + y^2/64 = 1

1. x is the major axis

2. y is the minor axis

3. sqrt 100 = +/-10 = (10,0) (-10,0)

4. sqrt 64 = +/-8 = (0,8) (0,-8)

5. 2 times sqrt 100 = 20

6. 2 sqrt 64 = 16

7. 64 = 100 – f^2

sqrt -36 = sqrt -f^2

f = +/-6 = (6,0) (-6,0)

Sketch if necessary.

The McDonald's Logo (Parabolas)

Pretty much two parabolas side-by-side. It would be a -x^2, that will be explained at a later time.

So many people did ellipses. O.o Ehhhh.....

Parabolas have only one quadratic, be it x^2 or y^2. (y=1/4x^2, x=1/4x^2).

If the equation is in standard form: y+h=a(x+k)^2 or x+h=a(y+k)2, you need to foil it out and solve for y or x.

Steps:

1) The first thing you can look at is the coefficient of the quadratic term. If it is a positive x^2, the parabola will open up. If it is a negative x^2, it will open down, like the McDonald's Logo! If it is a positive y^2, it will open to the right. If it is a negative y^2, it will open to the left.

2) The axis of symmetry is the line intersecting the vertex where if you “fold” the parabola in half, it would match up perfectly. The equation should be in the form y=ax^2+bx+c or x=ay^2+by+c. To find the axis of symmetry, you would find x=-b/2a or y=-b/2a. It is a line.

3) To find the vertex, you would take what x equals in the axis of symmetry for your x-coordinate, and plug that number into the function to find your y-coordinate. Vertex: (-b/2a, f(-b/2a)) The opposite would be true for y^2. Vertex: (f(-b/2a), -b/2a)

To find the focus and dirextrix, you need to find p using the formula 1/4p = your leading coefficient.

4) Your focus is a point.

For x^2, the x-coordinate stays the same as your vertex. Your y-coordinate is the vertex's y-coordinate + p.

If it is y^2, the y coordinate would stay the same and you would add p to the x-coordinate.

5) If it is x^2, y = the number you get from subtracting p from your vertex's y-coordinate.

If it is y^2, it would be x = the number you get from subtracting p from your vertex's x-coordinate.

Ex. 1

Sketch the graph of y=1/8x^2.

1. Opens up because it is positive.

2. Axis of symmetry: x=-b/2a
x=-0/2(1/8)
x=0
b is zero because there is no linear value.

3. Vertex: (0,(1/8*0^2) x-coordinate is 0, plug into y=1/8x^2 to get your y.
(0,0)

1/4(p) = 1/8, so p=2.

4. x-coordinate stays the same, add p=2 to y-coordinate.
Focus: (0,2)

5. Directrix: y=0-2
y=-2

To finish graphing, graph on the vertex a parabola opening up. Graph the axis of symmetry as a dashed line at x=0. Plot the focus at (0,2). Draw a dotted line for the directrix at y=-2. You directrix and focus should be equidistant from your vertex.

Ellipses

Standard form of an ellipse is x^2/a^2+y^2/b^2=1.

To find the equation of an ellipse, use these 7 steps.

1. First find which variable is the major axis. This will be the one with the largest denominator.

2. The minor axis is the variable with the smallest denominator.

3. Next find the vertex(the square root of the largest denominator). When putting this in point form, if the major axis is x, put the vertex in the x spot, and vice versa.

4. The other intercept is the square root of the smallest denominator. Put this point in the spot of the minor axis.

5. The length of the major axis is 2 times the square root of the largest denominator.

6. The minor axis' length is 2 times the square root of the smallest denominator.

7. To find the focus, use this formula: smallest denominator=largest denominator+focus^2.

_______________________________________________

Example problem:

Sketch the equation. x^2/100+y^2/64=1.

1. x is the major axis because 100 > 64.

2. y is the minor axis because 64 < 100.

3. The vertex is + or - 10 because that's the square root of 100. When putting this in point form, put the 10 and -10 in the x spot (10,0), (-10,0) because x is the major axis.

4. The other intercept is + or - 8. (0,8), (0,-8).

5. The length of the major axis is 2 times the square root of 100, which is 20.

6. The length of the minor axis is 16.

7. The focus is + or -6. (6,0), (-6,0).
64=100-f^2
-100 -100

(-f^2=-36)* -1

sqrt f^2= sqrt 36

f=+ or -6.

Ellipses :)

This week in class we learned about Ellipses.

The standard form for an Ellipse is x^2/a^2 + y^2/b^2 = 1.

To put in standard form you either divide or complete the square.

Step 1: The major axis is the variable with the largest denominator.

Step 2: The minor axis is the variable with the smallest denominator.

Step 3: The vertex is the square root of the largest denominator. To put in point form, if the major axis is x the answer goes in the x spot and vice versa.

Step 4: The other intercept is the square root of the smallest denominator. Then put in point form.

Step 5: The length of the major axis is 2 times the square root of largest denominator.

Step 6: The length of the minor axis is 2 times the square root of the smallest denominator.

-The focus is ALWAYS on the major axis and is a POINT.

Step 7: To find the focus, vertex, or the other intercept use the formulas below.

-Smallest denominator = largest denominator – focus^2

-Other intercept^2 = vertex^2 – focus^2

Example:

1.) x^2/4 + y^2/9 = 1

step 1: y

step 2: x

step 3: +/- 3 (0,3) (0,-3)

step 4: +/- 2 (2,0) (-2,0)

step 5: 2 X square root of 9 = 6

step 6: 2 X square root of 4 = 4

step 7: small deno = large deno – f^2

4 = 9 – f^2

-9 -9

-5 = -f^2

f^2 = 5

f = square root of 5

(0, square root of 5) (0, -square root of 5)

THE END :)

Ellipses!

Standard Form: x^2/a^2+y^2/b^2=1
--To put in standard form, either divide or complete the square.

Sketching an Ellipse:

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


2. Find the minor axis (the minor axis is the variable with the smallest denominator).


3. Find the vertex (the vertex is the square root of the largest denominator - to put in point form if the major axis is "x" the answer goes in the "x" spot and vise versa).


4. Find the other intercept (the other intercept is the square root of the smallest denominator, put in point form).


5. Find the length of the major axis (two times the square root of the largest denominator).


6. Find the length of the minor axis (two times the square root of the smallest denominator).


7. To find the foces, vertex, or other intercepts, use the formula: "smallest denominator=largest denominator-focus^2" or "other intercept^2=vertex^2-focus^2".

-- The focus is always on the major axis and is a point.

Example:

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

1. The "y" axis is the major axis because it has the largest denominator.


2. The "x" axis is the minor axis because it has the smallest denominator.


3. The square root of 9=+/-3, which gives you the vertices of (0,3) and (0,-3).


4. The square root of 4=+/-2, which gives you the other intercepts of (2,0) and (-2,0).


5. Two times the square root of 9(largest denom.)=2(3)=6


6. Two times the square root of 4(smallest denom.)=2(2)=4


7. Focus: 4=9-focus^2

-5=-f^2

Focus= +/- square root of 5

(0, square root of 5) and (0, -square root of 5)

Elipses

The standard form of an elipse is: x^2/a^2+y^2/B^2=1

*When you put in standard form, either divide or complete the square.


When you sketch an elipse:



1. Find the major axis. The major axis is the variable with the largest denominator.

2. Find the minor axis. The minor axis is the variable with the smallest denominator.

3. Find the vertices. In order to find the vertex, you find the square root of the largest denominator.

**To put in standard form, if the major axis is "x", the answer goes in the ''x'' spot, and vice versa.

4. Now you need to find the other intercept. To find this, you must square the smallest denominator.

***Put this into point form. (#,#)

5. Find the length of the minor axis by using two times the square root of the smallest denominator.

6. Find the length of the major axis by using two times the square root of the largest denominator.

7. Now you need to find the focus. The focus is always on the major axis and is always a point.

-To find the vertex, focus, or other intercept use:
smallest denominator=largest denominator-focus ^2
or
other intercept^2=vertex^2-focus^2



Example 1:

(x^2/36)-(y^2/16)

1. Major axis-X

2. Minor axis-Y

3. Square root of 36= 6

4. Square root of 16= 4

5. 2 times the square root of 36=12

6. 2 times the square root of 16= 8

7. 52=36-f2
f2=16

Ellipses

Standard Form of an Ellipse: (x^2/a^2) + (y^2/b^2) = 1
To Sketch an Ellipse:

1. Find the major axis. The major axis is the variable with the bigger denominator. The major axis is the longer part of the ellipse.




2. Find the minor axis. The minor axis is the variable with the smaller denominator. The minor axis is the shorter part of the ellipse.




3. Now you need the vertices. The vertices are on the major axis. It is the square root of the larger denominator in point form.




4. Find the other intercepts. The other intercepts are on the minor axis. It is the square root of the smaller denominator in point form.




5. Length of the major axis. The length of the major axis is the square root of the larger denominator times two.




6. Length of the minor axis. The length of the minor axis is the square root of the smaller denominator times two.




7. Find the foci. The foci are on the major axis. To find the foci you use the formula: (focus)^2 = (larger denominator) - (smaller denominator). Then put the foci in point form.

Example: Sketch the Ellipse

(x^2/16) + (y^2/25) = 1

1. The major axis is the y-axis because its denominator is larger.




2. The minor axis is the x-axis because its denominator is smaller.




3. The square root of 25 is +/- 5 so the vertices are (0,5) and (0,-5).




4. The square root of 16 is +/- 4 so the other intercepts are (4,0) and(-4,0)




5. The square root of 25 is 5 times 2 is 10 so the length of the major axis is 10 units.




6. The square root of 16 is 4 times 2 is 8 so the lenght of the minor axis is 8 units.




7. (focus)^2 = 25 - 16

(focus)^2 = 9

focus = +/- 3

so the foci are (0,3) and (0,-3)

This is something we learned in class to shorten time when you find the sketch of a graph. If the graph its self is not necessary using the discriminant formula will save you time. You must be warned though, it is not a complete short cut, it will only tell you what kind of graph it is.

*Your graph will not need to be in standard form (yay!). This formula will give you the shape of the graph : B^2-4AC. You identify the ABC in the equation and just simply plug into formula. Heres an example of how to do that
x^2+3xy+x^2=4
A=1 B=3 C=2

*Then take those numbers and plug them into an equation. Once you do that you will have to solve it for a single answer.
3^2-4(1)(2)
Simplify
9-8=1
Your answer is 1

*Then aplly your answer to these rules:
~If it is a circle your answer will be a -# and A=C & B=0
~If it is an ellipse your answer will be a -# and A wont =C & B wont =0
~If it is a parabola your answer will be 0
~If it is a hyperbola your answer will be +

*Take your answer and see what it is. The example is a hyperbola.

THE END

did you get this??

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