Click on the link below to access the assessment for conic sections. Remember that you can not go back once you hit submit. You have until Saturday, 5/2/20 to complete it.
Good luck! Next week we start the unit on matrices! Stay healthy everybody. docs.google.com/forms/d/e/1FAIpQLScp6fZ8cUqcPf3iI0FU3JTqgoXxkEM-lf1T0xvSnU5fohyOCA/viewform Since a number of people have asked, I am posting a review sheet below for practice on conic sections and will move the test to this Friday, 5/1/20. You will have until the end of the day on Saturday to take and complete the test. Stay healthy everybody!
Reminder - The final conic section homework on hyperbolas should be completed today. As always, if you have questions, email me. An assessment on this unit will be posted on Thursday, 4/30/20 and due by Friday, 5/1/20.
Remember, once you submit it, you can not go back and re-take it. Next week we will be starting the next unit which is Matrices. Stay healthy everyone! The definition of a hyperbola is the set of all points in a plane, the DIFFERENCE of whose distance from two fixed points (foci) is constant. The definition is similar to an ellipse, but instead of adding the distances from a point on the hyperbola to each of the two foci, we subtract those distances. For any point that you pick on the hyperbola, when you take the two distances from that point to each of the foci and subtract them, you will get the same value. The general format of the equation for a hyperbola is similar to that of an ellipse. The difference is that you have a subtraction sign between the x squared and y squared terms. The general form of the equation of a hyperbola centered at (0,0) is below. Unlike an ellipse, where the orientation was determined based on whether the bigger denominator was under the x or the y, with a hyperbola, the orientation is determined based on which coordinate is being subtracted, the x or the y. Image courtesy of www.mathwarehouse.com/hyperbola/graph-equation-of-a-hyperbola.php For the hyperbola with the horizontal transverse axis (y^2 is being subtracted), the basic properties are as follows: Vertices are at (a,0) and (-a,0) Equations of asymptotes are y= - b/a and y = b/a (slope m= (change in y) / (change in x) Note In this case the b is under the y and the a is under the x which is why we use b/a (y/x). Foci are at (c,0) and (-c,0) where c is defined as c^2 = a^2 + b^2. For a parabola with a vertical transverse axis (x^2 is being subtracted), you are pretty much swapping the coordinates. Vertices are at (0,a) and (0,-a) Equations of asymptotes are y= - a/b and y = a/b (slope m= (change in y) / (change in x) Note In this case the a is under the y and the b is under the x which is why we use a/b (y/x). Foci are at (0,c) and (0,-c) where c is defined as c^2 = a^2 + b^2. If you have a horizontal and/or vertical shifts, the format of the equations looks as follows: Note that to find the asymptotes for a hyperbola with a shifted center, use the point slope formula with (h,k) as your point, and the slope still being +/- ( a/b) or (b/a) depending on the orientation of the hyperbola. Included at the end of this post are some more notes on hyperbolas with examples worked out, the homework answers worked out and some good websites with videos on hyperbolas. Videos - basics of a hyperbola www.youtube.com/watch?v=i6vM82SNAUk Part I www.youtube.com/watch?v=6Xahrwp6LkI Part II Graphing hyperbolas centered at the origin. www.youtube.com/watch?v=W0IXOdsna9A&feature=youtu.be www.youtube.com/watch?v=vT-gnaAG51o&feature=youtu.be Graphing a hyperbola not centered at the origin. www.youtube.com/watch?v=_ssQfu8N6XE&feature=youtu.be www.youtube.com/watch?v=0qAB_5v_2G4&feature=youtu.be Finding the equation of a hyperbola given center, focus, vertex www.youtube.com/watch?v=ZwqNou_Z7oI&feature=youtu.be Not a video site, but goes into nice detail on how the calculations are derived. www.ck12.org/book/ck-12-algebra-ii-with-trigonometry-concepts/section/10.7/ www.ck12.org/book/ck-12-algebra-ii-with-trigonometry-concepts/section/10.8/ www.ck12.org/book/ck-12-algebra-ii-with-trigonometry-concepts/section/10.9/ Homework Assignment: pages 768-69: 2, 4, 6, 16, 18, 20, 28 pages 781-82: 10, 12 Due Tuesday - 4/28/20 There will be an assessment on this unit near the end of next week.
Geometric definition of an ellipse - The set of all points in a plane whose sum of distances from 2 fixed points called foci (plural of focus) is constant. The shape of an ellipse is like a circle that got squashed so it resembles a football. Inside the ellipse are two fixed points called foci. Pick any point on the ellipse and find the distance from it to focus 1, call it d1. From that same point, find the distance from it to focus 2, call it d2. The sum of d1 + d2 will always add to the same number no matter what point you chose to pick. Just like with the parabolas, these problems rely on knowing the general format of the equations and applying them based on the given information. x^2/a^2 + y^2/b^2 = 1. Equation of an ellipse centered at the origin. The general formats for the equations of ellipses centered at the origin and with shifts are in the word document below. The orientation of the ellipse, whether elongated more along the vertical axis or the horizontal axis is determined by which is the bigger denominator, a^2 or b^2. If the bigger denominator is under the x, then the main axis (longest) is the horizontal. If the biggest denominator is under the y, then the main axis (longest) is the vertical. So when graphing, the important thing is which coordinate is over the larger denominator, x or y. Below are some links showing the basic graph of an ellipse, and to solve some problems. Also included are notes with problems worked out and the answer key with all the homework problems worked out in detail. As always, please email me with any specific questions. Some time by the end of this week or early next week, you will have a short assessment on these problems. Homework due Friday: pages 759-60: 2, 4, 6, 10, 14, 20, 30 pages 781-82: 2, 4, 16 ( Hint for when you have difficulty getting the coefficient in front of the x^2 or y^2 to 1 when the right side is already 1 e:g: 9x^2 / 1 + 10y^2 / 1 = 1) Change the coefficient in front of the x^2 or y^2 to 1 and represent the denominator as 1/Coefficient of what had been in front of the x or y. The above equation is equal to: x^2 / (1/9) + y^2 / 1/10 = 1 This is now in standard form. Link for ellipses. Basic concepts. Parts 1 and 2 www.youtube.com/watch?v=LVumLCx3fQo www.youtube.com/watch?v=oZB69DY0q9A Graphing ellipses centered at the origin with major axis along x and along y www.youtube.com/watch?v=azI5kALyiXs&feature=youtu.be (horizontal axis) www.youtube.com/watch?v=3qckea8OuN8&feature=youtu.be (major axis vertical) Graphs not centered at the origin: www.youtube.com/watch?v=tJmp1PJD9o8&feature=youtu.be www.youtube.com/watch?v=oWGyVpq94CM&feature=youtu.be Given an equation, rewrite in standard form and graph. www.youtube.com/watch?v=-i48L0WQ-2I&feature=youtu.be www.youtube.com/watch?v=CL7SNu-5riw&feature=youtu.be Given some information about the ellipse (e.g focus, vertex, etc), find the equation. www.youtube.com/watch?v=k5gMLDjMfvI&feature=youtu.be www.youtube.com/watch?v=RWaEIJOlHlw&feature=youtu.be There are a number of videos on this website for ellipses: www.mathispower4u.com/alg-2.php in addition to the ones posted above.
This unit takes a geometric approach to parabolas, ellipses and hyperbolas. For this unit, you need to understand the general format of the equations for each of these conic sections. Getting the equation in general format will also often require the technique of completing the square. The first conic section to be discussed will be the parabola. The two parts that define the path of a parabola are the Focus ( a point) and the directrix (a line). The definition of a parabola is the set of all points equally distant from a focus and the directrix. If some point (x,y) on a given parabola is a distance of 5 from the focus, it is also 5 from the directrix. If another point(x2,y2) is a distance of 10 from the focus on the this same parabola, it is also a distance of 10 from the directrix. Every point on the parabola meets this requirement. The focus will be somewhere on the axis of symmetry of the parabola (the line that bisects the parabola). The directrix line and the focus point are on the opposite sides of the turning point or vertex of the parabola and the vertex will be halfway between the focus and directrix. General format of equations: Given a parabola with a vertical axis of symmetry. (points up and down) and a vertex at (0,0) Vertex = (0, 0) Focus point is (0,P) (P = focal point, is on the y axis) Directrix is the equation y = - p General format of the equation is: x^2 = 4py Given a parabola with a horizontal axis of symmetry (points left and right) and a vertex at (0,0) Vertex = (0,0) Focus point is (P,0) (P = focal point on the x axis) Directrix is the equation x = -p General format of the equation is y^2 = 4px If the parabola has been shifted either horizontally, vertically or both, the general format of the equations are: vertical parabola = ( x-h)^2 = 4p(y-k) where the vertex has been shifted to the point (h, k) Note how the sign of h and k change when going between equation and vertex. horizontal parabola = (y - k)^2 = 4p(x-h) where the vertex has been shifted to the point (h,k) Note in the formulas above that: The x coordinate of the vertex, (h) is always with the x in the equation The y coordinate of the vertex ,(k) is always with the y in the equation. When trying to find the vertex, focus and directrix of a parabola that does not have its vertex at the origin (0,0), it is often helps to figure out what these would be without the shift and then applying the shift. For example: Equation = (x-2)^2 = 12(y-3) This is the formula for a parabola that is vertical The parenthesis tell us the shift is 2 right on the x axis and 3 up on the y (Shift is always in the opposite direction of the sign) Shift = (2,3) Totally ignore the shift in parenthesis and the equation looks like x^2 = 12 y This is the general form of x^2 = 4p y 1) Solve for p. 4p = 12 in this example (The coefficient in front of y always = 4p). So p must = 3. Therefore from the formulas above: Pre - shift Shift(2,3) Post Shift Vertex = (0,0) x+2, y+3 Vertex = (2,3) Focal point = (0,3) Focal Point = (2,6) Directrix = y= - 3 Directrix y = 0 (added 3 to the y from the shift) Attached below are notes explaining in more detail and examples worked out as well as some helpful links. The answers to the homework questions are posted again as well. Homework problems: p. 751 #'s 2,4,6,12,28,40 pp. 781-782 #'s 6,8,14,22 Due Tuesday 4/21/20 General description of various conic sections. www.youtube.com/watch?v=iJOcn9C9y4w Description of parabola conic section and basic equations www.youtube.com/watch?v=k7wSPisQQYs Parabola part II www.youtube.com/watch?v=CKepZr52G6Y There are a number of specific examples on this page for graphing parabolas with and without shifts. Once you click on the link below, click Ctrl f and a search box opens up. Type parabola and you will be taken to the section of the page with many different videos. www.mathispower4u.com/alg-2.php As always, email me with any questions.
Good luck. Click on the link for the first assessment for the 4th quarter.
docs.google.com/forms/d/e/1FAIpQLScD4MKe5lJEXZ3StxqvQNmqirfiDjWox7hk5CK7qCAwRS-ZvA/viewform The last type of equations to review are parametric equations. A parametric equation is similar to the normal rectangular equations with x's and y's, except they have an extra parameter (variable) for time - (t). It is generally used to track the course of an object along some kind of path (this is an equation in terms of x and y coordinates) but the t tells where the object is located on the path at a give time t: - Think of tracking the path of an asteroid or comet as it passes through the solar system. The entire track of the comet is generally described by the equation of an ellipse, but the t component tells where the comet is on this elliptical path at time t. When converting from parametric to rectangular equations, you want to be able to get rid of all the t's in the equations. The main techniques are to either 1) Rewrite one of the equations to get t by itself on one side with all the x's and y's on the other. Then then use that value for t in terms of x and y as a substitute in the second equation to replace all ts. 2) If you have sin t and cos t and you want to get a t by itself, you will often need to write in terms of the identity (sin t)^2 + (cos t)^2 = 1. And then do a suitable substitution for t. Some helpful videos: Graphing parametric equations with the TI 84 www.youtube.com/watch?v=4Y14XhPD7Os Converting parametric equations to rectangular form www.youtube.com/watch?v=tW6N7DFTvrM Converting parametric equations of a line into a rectangular equation (problem 24) www.youtube.com/watch?v=BGWX0TOthsY&feature=youtu.be Included below is a pdf file with some notes showing examples worked out, a pdf file of the review sheet for the entire unit, and the pdf files of the answer keys for the homework and review sheet. Your homework from the book pages is: Page .807: 2, 5, 6, 8, 11, 14, 16, 24 Problem 24 is just like one of the problems explained in the class notes included. Your second part of the homework is the review sheet listed below. Both assignments should be completed by Thursday, 4/16/20. Your FIRST ASSESSMENT for the 4th quarter will be due at the end of Friday 4/17/20. The link will be posted Friday morning by 8:00 a.m. You will have until midnight 12:00 a.m. on Saturday, giving you a 16 hour window in which to complete it. After that, the assessment will be locked. Please note that once you submit this exam, you will not be allowed back in to submit it again. As always, email me with any questions.
As a continuation from the previous assignment, you are to try more practice using the TI-84 to graph the various polar equations and convert from rectangular to polar form. The video link showing graphing using the TI-84 was: www.youtube.com/watch?v=PZwiiZQhM0c&feature=youtu.be Do Problems pp. 594-95 from the problem packet document previously posted. (Button on right side of page) Verify that you correctly graphed problems 2,4,6. Also graph problems 16, 20, 22, 24, 26.
Due 4/9/20 Polar equations - 1) page. 587 -(problem pages listed on previous post - NOT IN TEXTBOOK). Problem numbers 42-54 evens and number 58 2) Also look at page 594 on the problem page file. It lists the general shapes and generic form of equations for various polar curves. Practice plotting some of the equations and try changing around the values for a, b and r in the different equations to see the effect on the graphs in your calculator. Graph problems 1-6 at the bottom of page 594 - Exercises section 8.2. Make sure you have the proper settings on your calculator, radian, polar, etc. Be aware of the settings of your maximum and minimum x and y values on your calculator as well. See the video below. 3) If you have not already done so, take the "quiz" that was assigned by posting on the link below. https://sites.google.com/sachem.edu/mr-ks-math-and-computer-land/math-12-precalculus The last question on the quiz is to enter the period for your class. While it is now late, you will get some credit it it is taken by the end of tomorrow. Attached below are the notes for this section, the answer key to the homework on page 587 and some links that demonstrate several examples of converting equations between polar and rectangular as well as how to graph an equation in polar form on your TI-84. Read the notes and look at the video examples first. Some review: Video showing how to convert from (x,y) rectangular coordinates to the equivalent in polar coordinates (r, theta). https://www.youtube.com/watch?v=Vg88CWOlDTY Graphing polar equations in the TI 84 https://www.youtube.com/watch?v=PZwiiZQhM0c&feature=youtu.be Example of a conversion from rectangular (circle) to polar form. https://www.youtube.com/watch?v=tUH6tUiJbH8&feature=youtu.be Example converting a parabola in rectangular form to polar form. https://www.youtube.com/watch?v=XkZAtV8jNb4&feature=youtu.be Video showing how to convert equations from polar back to rectangular coordinates form. https://www.youtube.com/watch?v=IKbRiU7kL2w Extra notes: Essentially you are doing identities where you are converting the trig into polar form or vice versa. To convert from polar to rectangular you want to get rid of all the r's and theta's in the equation and change to terms of x and y. To summarize: There are three main types of techniques as listed below to convert between polar and rectangular forms: 1) You can multiply both sides by r to get: r squared - converts to x squared + y squared or to get: r cos theta - converts to x or to get: r sin theta - converts to y 2) Sometimes you only have r on 1 side of the equation and you don't want to multiply r to the other side because it is already in x and y terms. In this case, SQUARE both sides of the equation so that one side becomes r squared which is now = x squared + y squared. The other side is already in x and y terms, but now squared. So it is still in rectangular form. 3) If you have a theta by itself on one side, take the tangent of both sides of the equation. Tan(theta) = Tan(right side) Since tan x = sin x/cos x (Identity), which is exactly the same as r sin x / r cos x, the left side can now be written as y/x. Then finish the algebra to solve the equation. Email me with your questions.
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