Chapter+12

//**Chapter 12: Analytic Geometry**// toc

12.1 Conics
media type="youtube" key="oI51LBUYR0c" height="385" width="480" Conics are equations that can all be represented by two nappes in conic meeting at a single point. The equation itself can be thought of as a flat plane that can be slid through these nappes at different angles to make different parts overlap. The parts that overlap are what we see as an equation. Depending on the angle of the plane, either a circle, ellipse, parabola, or a hyperbola can be formed.



12.2 The Parabola
**Vertex at (** //**h,k** // **)** The vertex of a parabola isn’t always at //(0,0)// or any conic for that matter. The variable //h// is the no. of units shifted horizontally and the variable //k// is the no. of units shifted vertically.


 * The equation for a parabola shifted __horizontally__ is: **// (y - k)^2=__+__4a(x - h) //**
 * It’s ** “+4//a//” ** if it opens ** right ** and ** “-4//a//” ** if it opens ** left **.
 * The equation for a parabola shifted __vertically__ is: **// (x - k)^2=__+__4a(y - k) //**
 * **// It’s // “+4a” // if it opens up and “-4a” if it opens down .//**

** Finding the Vertex, Focus, Directrix, and Latus Rectum (L.R.) of a Parabola and Graphing a Parabola ** (y-2)^2=8(x+1)

**Finding the Equation of a Parabola from a graph**

Knowing that the vertex is at (-2,0), a point on the graph is at (0,1), and the parabola is opening right, we get this info: First, try it by yourself. Then, highlight the box below for the correct method and solution.

(y-k)^2=c(x-h) (y-0)^2=c(x-(-2)) y^2=c(x+2) Plug in the point (x,y) into the equation 1^2=c(0+2) 1=2c c=1/2 y^2=1/2(x+2) **Finding the equation of a parabola and graphing it using the vertex and directrix ** __V: (1/2,0) D:x=-1/2__ a=1 Opens to the right ((y-k)^2=4a(x-h)) Focus: (3/2,0) Latus Rectum=4 (y-0)^2=4(1)(x-1/2) Equation: y^2=4(x-1/2)

**Finding the equation of a parabola and graphing it using the vertex, axis of symmetry(A.O.S.), and a point on the graph __V: (0,0) A.O.S.:y-axis P: (2,3)__ ** Opens up (x^2=4ay) (2)^2=4a(3) 4=12a a=1/3 F: (0,1/3) D: y=-1/3 Latus Rectum: 4/3 *4a=4(1/3)=4/3* Equation: x^2=4/3y ** Completing the square to get the standard equation of a parabola and graph it __x^2+8x=4y-8__ ** (x^2+8x )=4y-8 Complete the square  1 (x^2+8x +16 )=4y-8+( 16 X 1 ) Equation: (x+4)^2=4(y+2) Opens up a=1 V:(-4,-2) F:(-4,-1) D:y=-3 L.R.: 4
 * Plug in the point (x,y)

<span style="font-family: arial,helvetica,sans-serif; line-height: 20px;"> <span style="display: block; font-family: arial,helvetica,sans-serif; font-size: 110%; line-height: 20px; text-align: left;">Here are some links that also explain parabolas: <span style="color: #000000; font-family: 'Courier New',Courier,monospace;"> [] [] <(look at the 2nd question) <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 110%;"> Practice Problems: <span style="color: #000000; font-family: 'Courier New',Courier,monospace;"> [] A Paraboloid of Revolution is a parabola spinning around on its axis of symmetry forming a 3D shape; a 3D parabola. Examples of these are satellite dishes, searchlights, the reflector of a flashlight, and the mirror in a telescope. In math, the vertex is normally at (0,0) and the image is a parabola opening up. The equation //** x^2=4ay **// is used in these cases. The equation //** x^2= -4ay **// can be used for P.R.'s facing down.
 * <span style="color: #ff0000; font-family: 'Courier New',Courier,monospace; font-size: 150%;">Paraboloid of Revolution **

Ex. A flashlight’s reflector is a paraboloid of revolution. If its diameter is 6 inches and its depth is 2 inches, how far from the vertex does the light bulb need to be put? Using the equation x^2=4ay, do the following steps to get the answer.



= =

<span style="background-color: #000080; color: #19f01d; display: block; font-size: 120%; text-align: center;">12.3 The Ellipse
In this section you will be finding the equation of an ellipse, graphing ellipses, discussing the equation of ellipses, and working with ellipses with the center at (h,k) or not at origin. An ellipse is the collection of all points in the plane the sum of whose distances from two fixed points, called the foci, is a constant. The vertex of an ellipse isn’t always at // (0,0) //  or any other conic for that matter. The variable // h //  is the number of units shifted horizontally and the variable  // k //  is the number of units shifted vertically. The line containing the foci is called the major axis, and the midpoint of the line segment joining the foci is called the center of the ellipse. Every ellipse comes with 2 points called the vertices too. They intersect through the major axis and the ellipse, and the distance from 1 vertex to the other is called the length of the major axis. An equation of an ellipse with the center at origin, and foci at ( ±<span style="font-family: arial,helvetica,sans-serif; line-height: 20px;"> c,0) and vertices at ( ±<span style="font-family: arial,helvetica,sans-serif; line-height: 20px;">a,0) is

<span style="display: block; font-family: Arial,Helvetica,sans-serif; font-size: 20px; text-align: center;">x^2/a^2+y^2/b^2=1 where a>b>0 and b^2=a^2-c^2. The major axis is the x-axis.

An equation of an ellipse with the center at origin, but foci at (0, ±c) and vertices at (0, ±<span style="font-family: arial,helvetica,sans-serif; line-height: 20px;">a) is x^2/b^2+y^2/a^2=1 where a>b>0 and b^2=a^2-c^2.

<span style="display: block; font-family: Arial,sans-serif; font-size: 10pt; line-height: 115%; text-align: center;">**Example 1.** To find an equation of an ellipse with center at origin, one focus at (2,0), and a vertex at (-3,0) you must use b^2=a^2-c^2, where a is the distance from the vertex to the center, and c is the distance from the foci to the center. So it would turn out to be, b^2=9-4, therefore b^2=5
 * Finding the equation of an Ellipse **

**The equation for this ellipse is x^2/9+y^2/5=1**


 * <span style="display: block; font-family: Arial,sans-serif; font-size: 10pt; line-height: 115%; text-align: center;">Example 2. **

To find the equation of an ellipse with the center at origin, one focus at (0,-4), and a vertex at (0,5) you also use the same formula b^2=a^2-c^2, but a is 5 because the major axis is on the y-axis. So b^2=25-16, therefore b^2=9
 * The equation for this ellipse is x^2/9+y^2/25=1**
 * Finding and discussing the equation of an Ellipse <span style="font-family: Arial,sans-serif; font-size: 10pt; line-height: 115%;">Example 1. Find the equation: **

The equation is that of an ellipse, center at origin, and major axis along the x-axis, with a^2=16, and b^2=9, and to find its foci use c^2=a^2-b^2 which is the same as b^2=a^2-c^2 just added c^2 to the both sides and subtracted b^2 from both sides which makes c^2=16-4, c^2=√12 or 2√3 Foci are at ( ±2√3,0)<span style="font-family: arial,helvetica,sans-serif; line-height: 20px;"> and vertices at ( ±4,0)



**The equation for this ellipse is x^2/16+y^2/4=1**

First you must put it into standard form, which is x^2/1+y^2/4=1, by dividing both sides by 36. The larger number, 9, is in the denominator of y^2, so this equation is of an ellipse with the center at origin, and major axis along the y-axis. Therefore a^2=4, b^2=1, and c^2=4- 1=3. The vertices are at (0, ±2 ) and the foci are at (0, ± √3) To graph an ellipse using a graphing utility, first you must solve for y. Subtract both sides by x^2/16: y^2/9=1-x^2/16 Multiply both sides by 9: y^2=9(1-x^2/16) Take the square root of both sides: y= ±√9(1-x^2/16) You're done, but on the graphing utility, you must use both the positive, and negative equation for the ellipse to appear.
 * Example 2. Discuss the equation: 36x^2+9y^2=36 **
 * Graphing Ellipses **
 * Example 1. Graph x^2/16+y^2/9=1 **

**Center at (h,k)** If an ellipse with center at the origin and major axis corresponding with a coordinate axis is shifted horizontally h units and then vertically k units, it will give you an ellipse with center at (h,k) and major axis parallel to a coordinate axis. In such a situation, x is replaced with x-h, and y is replaced with y-k. An ellipse that is parallel to the x-axis, foci at (h ± c,k), and vertices at (h ±a ,k) has the equation: An ellipse that is parallel to the y-axis, foci at (h, k±c), and vertices at (h, k±a) has the equation: If the center of an ellipse is at (-3,4), and has a focus at (-3,6), and a vertex at (-3,7), give the equation. First you know the center is at -3,4 so to start off the equation it would be (x+3)^2/b^2+(y-4)^2/a^2=1,because the the major axis is the y-axis, now solve for a, b and c. because B^2=A^2-C^2, B^2=9-4, therefore b^2=5 and the equation is (x+3)^2/5+(y-4)^2/9=1 and the vertices are (-3,7) and (-3,1), and foci are (-3,6) and (-3,2).
 * (x-h)^2/a^2+(y-k)^2/b^2=1**
 * (x-h)^2/b^2+(y-k)^2/a^2=1**
 * Finding the equation of an Ellipse, center not at origin **

**Completing the square for an Ellipse** Given the equation, x^2+4x+4y^2-8y+4=0, discuss the ellipse. First you must complete the square, factor x into (x+2)^2 and left with 4y^2-8y with can be factored into 4(y^2-2y); then add 4 to both sides to get 4(y-1)^2, and last divide both sides by to get (x+2)^2/4+(y-1)^2=1 To discuss the equation, you can find the center at (-2,1), and vertices at (-4,1), and (0,1). To find the foci use b^2=a^2-c^2, and you get 1=4-c^2, therefore c=√3 and the foci are (-2-√3,1) and (-2+√3,1).

<span style="background-color: #000080; color: #19f01d; display: block; font-size: 120%; text-align: center;">12.4 The Hyperbola
A hyperbola is the set of all points in the coordinate plane whose difference of of their distance to two fixed points (foci) is constant. Below are the parts of a hyperbola: There are three values used in defining a hyperbola: //a//, //b//, and //c//. //a// is the distance from the vertices to the center, //c// is the distance from the foci to the center, and //b// is related to wide the hyperbola is. //a, b,// and //c// are all related with the equation //a²+b²=c²,// so you can find the value of one from the other two.
 * __Definition:__ **

**__The equation and asymptotes:__** The equation for the hyperbola is //x²/a² - y²/b²//=1, where the hyperbola has its center at the origin, vertices //a// units away from the center on the transverse axis, and foci c (//c=√(a²+b²)// units away from the center on the transverse axis. In this scenario, the hyperbola's transverse axis is the x axis, as it contains the vertices and foci. If you have an equation //y²/b² - x²/a²=1//, (notice //y// comes first) then the transverse axis will be the y axis, so the vertices and foci will be on there too.

**__Graphing it:__** To graph a hyperbola, you must also find the asymptotes of it. They are the lines that go through the corner of the "box" as seen above. The box contains the point (±a,±b), so it follows that the asymptotes are lines passing through the center, with the slope ±//a/b,// and so the equations are //y= (a/b)x// and //(-a/b)x.//

Lets say we have this hyperbola. It has a center at the origin, vertex at (3,0), and a foci at (5,0). Find the equation and graph.

From the above information, we can tell the transverse axis is the x axis, as it contains a vertex and focus. It must also have another vertex at (-3,0) and focus at (-5,0). //a//=3, as the vertices are three away from the center. We know c is 5 becuse the foci are 5 away from the center. To find //b//, we use //b²=c////²-a////²,// and get //b=4//. Now that we know //a// and //b//, we can get the asymptotes: //y=(3/4)x,// and //y=(-3/4)x.//

The info: Transverse axis: x axis Center: (0,0) Vertices: (±3,0) Foci: (±5,0) Asymptotes: //y=±(3/4)x// a=3 b=4 c=5

We can now get the equation: //x²/9 + y²/16//=1

And the graph: **__Transforming Hyperbolas:__** Hyperbolas can be moved around by substituting x and y for (x-h) and (y-k) to make the center (h,k), just like with any equation. The substitution applies to everything-equations, points, asymptotes.

Lets say we want to find another hyperbola's equation and graph it. It has vertices at (-1,0) and (-1,4), and a focus at (-1,-2). First off, it has a vertical transverse axis, because the vertices are on a vertical line. We know the center is halfway between the vertices, so it has to be at (-1,2). The vertices are two away from it, so //a=2//. The other focus is four away from the center, like the first, but in the other direction, so it is at (-1,6). Also, //c=4//. We can find //b,// it is √(//c//²-a²), which is √12, or approximately 3.4. Armed with the knowledge of //b//, we can find the asymptotes: //(y-2)=(±2/√12)(x//+1). (Remember that we need to transform the asymptotes too!)

The info: Transverse axis: x=-1 Center: (-1,2) Vertices: (-1,0) and (-1,4) Foci: (-1,-2) and (-1,6) Asymptotes: //(y-2)=(±2/√12)(x//+1) a=2 b=√12 (~3.4) c=4

The equation: //(y-2)²/4 + (x+1)²/12=1//

The graph:

<span style="background-color: #000080; color: #19f01d; display: block; font-size: 120%; text-align: center;">12.5 Identifying Conics in General Form
A conic equation is in general form when it is in the form: //ax²+bxy+cy²+dx+ey+f=0// To identify the conic, you just need to pay attention to a and c.

If either a or c, but not both is 0 (x² or y² not present), it is a parabola. If a=c, and they are not zero, it is a circle. If a≠c, then it is an ellipse. If a and c are of opposite sign, then it is a hyperbola.

Examples:

//3y²+2x-y=0// Parabola

//-4x²+y²-6x+3=0// Hyperbola

//x²+y²-7x+2y-3=0// Circle

3x²+2y²-5x+4y-1=0 Ellipse

<span style="background-color: #000080; color: #19f01d; display: block; font-size: 120%; text-align: center;">12.8 Systems of Nonlinear Equations
<span style="font-family: Arial,Helvetica,sans-serif; font-size: 10pt; line-height: 115%;">Like in systems of linear equations, the solution(s) in a system of nonlinear equations is/are the point(s) of intersection of the graphs of the equations. Also like in systems of linear equations, there are multiple ways to find the solution to a nonlinear system. <span style="font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: 115%;">


 * <span style="color: #ff0000; font-family: Arial,Helvetica,sans-serif; font-size: 140%;">Graphing **

<span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 10pt; line-height: 115%;">Graphing is not the most reliable way of solving a system of nonlinear equations, as the points on a graph are merely an approximation of the exact value(s) of the solution(s). Recall how to find the shape of the graph of a given equation:
 * Straight line: //No squared values.//
 * Parabola: //One squared value.//
 * There are three for two squared values:


 * 1) Hyperbola: //Coefficients of the squared values are of opposite signs.//
 * 2) Ellipse: //Coefficients are of the same sign but are unequal.//
 * 3) Circle: //Coefficients are equal (sign and value).//

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 10pt;">One good reason to graph nonlinear equations in a system is to see the approximate number and location of the solution(s) of the system.

<span style="color: #0000ff; font-family: Arial,Helvetica,sans-serif; font-size: 120%;">Example 1: Solve by Graphing <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">The solutions appear to be at <span style="color: #19f01d; font-family: Arial,Helvetica,sans-serif;">(-1/2, 1/2) <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;"> and <span style="color: #19f01d; font-family: Arial,Helvetica,sans-serif;">(2, 8) <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">.


 * <span style="color: #ff0000; font-family: Arial,Helvetica,sans-serif; font-size: 140%;">Substitution **

Another method for solving a system of equations is substitution. In this method, a variable is solved for in one equation, and the resulting (and equivalent) equation is plugged into the other. We use the previous example:

<span style="color: #0000ff; font-family: Arial,Helvetica,sans-serif; font-size: 120%;">Example 2: Solve by Substitution


 * Elimination **

Sometimes, it's easiest to add two equations together to eliminate like terms. Other times, subtraction is useful, but I would think of subtraction as adding a negative. For example:

Example 3: Solve by Elimination (this problem, #31, can be found in your book on page 971)

Now, try one on your own: 3x^2-2y^2+5=0; 2x^2-y^2+2=0. Highlight for solution.

First, multiply both sides of the second equation by -2: 3x^2-2y^2+5=0; -4x^2+2y^2-4=0. Next, add both equations together: -x^2+1=0. Move the x^2 to the other side, take the square roots of both sides. x=+/- 1. Plug x into one of the equations (I used the first). y=+/- 2. Your solution set is (1, 2), (1, -2), (-1, 2), (-1, -2).

** Inconsistent Systems ** Sometimes there are no solutions to a system of equations--the graphs just never intersect.

Example 4: Solve.

Additional Resources: http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut52_nonlinear_sys.htm http://www.purplemath.com/modules/syseqgen3.htm For review over systems of equations: http://a2h-1st-hour.wikispaces.com/Chapter+7