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.
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