envelope Posted February 20, 2007 Share Posted February 20, 2007 The text book I borrowed from my school reads "you already know that the graph of y = ax^2 is a parabola whose vertex (0,0) lies on its axis of symmetry x=0." On the contrary, I know not of what it reads. I am unaware of what y symbolizes, as well as what a & x^2 symbolizes. The axis of symmetry as defined by the text book is "the line perpendicular to the parabola's directrix & passing through its focus. In particular, the axis of symmetry is the vertical line through the vertex of the graph of a quadratic function." So, from this, I've deduced that the axis of symmetry is the line perpendicular to a fixed line associated with the parabola, which passes through the focus (which aids in decsribing the conic section). & not only do parabolas have a vertical axis of symmetry, opening up or down, but a horizontal axis of symmetry, opening left or right. It then goes onto list & show each case: x^2 = 4py, p > 0 x^2 = 4py, p < 0 y^2 = 4px, p > 0 y^2 = 4px, p < 0 The top two being parabolas with a vertical axis of symmetry, opening up & down. The bottow two being parabolas with a horizontal axis of symmetry, opening left or right. What I don't understand is what the equation is stating. In particular, what x^2 is symbolizing, what 4py is symbolizing, & p > 0 is symbolizing, & so on & so forth throughout each case. Another thing I am not understanding are the standard equations of a parabola with its vertex as its origin. The equations being: x^2 = 4py, its focus (0,p), its directrix y = -p, and its axis of symmetry being vertical (x = 0) y^2 = 4px, its focus (p,0), its directrix x = -p, and its axis of symmetry being horizontal ( y = 0) I am guessing that in order for me to under the standard equations, I should first understand what each individual component means. Pertaining to graphing an equation of a parabola, I am having trouble grasping a few things taken from the text book. First off, it says identify the focus and directrix of the parabola given by x = -1/6y^2. In order to get rid of the fraction, I understand you multiply by -6 on each side. So: (-6)x = -1/6y^2(-6) giving you -6x = y^2. Because the variable is y^2 the axis of symmetry is horizontal. Now, the equation above, the text book says, will aid in finding the focus and directrix. It then says that "since 4p = -6, you know p = -3/2. The focus is (p,0) = (-3/2, 0) and the directrix is x = p = 3/2." It then goes onto saying that since p < 0, only negative x-values should be chosen to create a table of values. & then the table of values goes onto showing: x = -1, -2, -3, -4, -6 y = +-2.45, +-3.46, +-4.24, +-4.90, +-5.48 What I am not understanding is how the y values are arrived at. As you can probably already tell, I'm as lost as they get. If you haven't already labeled me mathematically deficient, I would greatly appreciate any help. In-depth explanations would be ideal. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted February 20, 2007 Share Posted February 20, 2007 The text book I borrowed from my school reads "you already know that the graph of y = ax^2 is a parabola whose vertex (0,0) lies on its axis of symmetry x=0." On the contrary, I know not of what it reads. I am unaware of what y symbolizes, as well as what a & x^2 symbolizes. The axis of symmetry as defined by the text book is "the line perpendicular to the parabola's directrix & passing through its focus. In particular, the axis of symmetry is the vertical line through the vertex of the graph of a quadratic function." So, from this, I've deduced that the axis of symmetry is the line perpendicular to a fixed line associated with the parabola, which passes through the focus (which aids in decsribing the conic section). & not only do parabolas have a vertical axis of symmetry, opening up or down, but a horizontal axis of symmetry, opening left or right. Right, you're going in the right direction. Let me explain further. If you are familiar with the Cartesian plane (x and y coordinate system), x is associated with the horizontal position of a point, and y with the vertical position. Thus, (1, 2) is a point 1 unit to the right and 2 units above the origin (center of the graph). When you have an equation such as [math]y = ax^2[/math], it is describing the relationship between those points. That means that for each point, the y value (vertical position) is the square of the horizontal position (x), multiplied by a (some arbitrary value; when you're given a real equation, a will be a number). The axis of symmetry of a parabola like this one: would be a line going vertical directly through its center. It's called a line of symmetry because if you were to divide the parabola down this line, each side would be symmetrical - totally identical, only flipped. Assuming that the center of that parabola is located where x = 0 on the plane (the center of the graph), the axis of symmetry is the graph of the line [math]x = 0[/math]. I'm afraid I don't know what some of the other equations you have are showing, but I can continue here: I am guessing that in order for me to under the standard equations, I should first understand what each individual component means. Pertaining to graphing an equation of a parabola, I am having trouble grasping a few things taken from the text book. First off, it says identify the focus and directrix of the parabola given by x = -1/6y^2. In order to get rid of the fraction, I understand you multiply by -6 on each side. So: (-6)x = -1/6y^2(-6) giving you -6x = y^2. Because the variable is y^2 the axis of symmetry is horizontal. You can shortcut that: if the equation is x = something, it will be horizontal; if the equation is y = something, it will be vertical. Here's a good link which links to other pages that describe everything: http://en.wikipedia.org/wiki/Parabola If you don't know what something is, type it in to the search box on the left. Don't worry about us labeling you mathematically deficient; it is apparent that you care enough about learning to come to us for help, and you are very good at explaining what you need help with. Unfortunately, I don't know some of the terms myself; I'm guessing I just learned parabolas differently. Good luck. Link to comment Share on other sites More sharing options...
JackMuChabas Posted August 17, 2007 Share Posted August 17, 2007 Problems with conical indices given a quantuum figure. Link to comment Share on other sites More sharing options...
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