Sullivan, 8th ed. MAC1140, MAC114711.2: The ParabolaThe vertex V is the midpoint of the segment from F to D: d(F, V) =d(V, D). If d(F, V) = a, then 2a are two other points on the parabola;the line connecting them is called the latus rectum. Equations ofparabolas with vertex (h, k) are (x-h)2 = 4a(y-k) if axis of symmetryis vertical and (y-k)2 = 4a(x-h) if axis of symmetry is horizontal. Find the equation of the parabola defined, find the two points that define the latus rectum, and graph the equation:1. F = (-4, 0); V = (0, 0) 2. D: y = -1; F = (0, 1)a __Equation L.R. a Equation L.R. 3. V = (0, 0); contains (-2, -4) 4. F = (-4, -2); D: y = 4Axis of Symmetry: x-axisa Equation L.R. a Equation L.R.Sullivan, 8th ed. MAC1140, MAC1147Find the vertex, focus, and directrix of each parabola; graph it.5. (y + 1)2 = -4(x – 2) 6. x2 + 6x – 4y + 1 = 0 Write an equation for each parabola:7. 8.9. The cables of a suspension bridge are in the shape of a parabola. The towers supporting the cables are 1000 feet apart and 135 feet high. If the cables are at a height of 10 feet midway between the towers, what is the height of the cable at a point 250 feet from the center of the bridge?Sullivan, 8th ed. MAC1140, MAC114710. A bridge is in the shape of a parabolic arch and has a span of 150 feet. The height of the arch60 feet from the center is 15 feet.. Find the height of the arch at its center.Sullivan, 8th ed. MAC1140, MAC1147Page 740 – 742:36. Find the vertex, focus, directrix, and latus rectum for x2 + 6x – 4y + 1 = 0; graph the equation:52. A cable-TV receiving dish is in the shape of a paraboloid of revolution. Find the location of the receiver, which is placed at the focus, if the dish is 6 feet across and 2 feet
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