Today in Precalculus• Go over homework• Notes: Parabolas

– Completing the square– Writing the equation given the graph– Applications

• Homework

Example 1Prove that the graph of y2 + 2y – 8x – 7 = 0 is a parabolay2 + 2y = 8x + 7 y2 + 2y + 1 = 8x + 7 + 1y2 + 2y + 1 = 8x + 8(y+1)2 = 8 (x + 1)Standard form for a parabola

Example 1(y+1)2 = 8 (x + 1)Graph:

Opens to the rightvertex (-1, -1)4p = 8, p = 2focal length =2focal width = 8focus: (1, -1)directrix: x = -3axis: y = -1

Example 2Prove that the graph of x2 – 2x + 3y + 7 = 0 is a parabolax2 – 2x = -3y – 7 x2 – 2x + 1 = -3y – 7 + 1x2 – 2x + 1 = -3y – 6(x – 1)2 = -3(y + 2)Standard form for a parabola

Example 2(x – 1)2 = -3(y + 2)Graph:

Opens downwardvertex (1, - 2)p = -3/4focal length = -3/4focal width = 3focus: (1, -11/4)directrix: y = -5/4axis: x = 1

Writing an equation of a parabolaFind vertex, substitute into general form vertex: (-2, -2)(y+2)2=4p(x+2)Find another point and substitute for x and y (1,-5) (-5 + 2)2 = 4p(1 + 2) Solve for p9 = 12pp = ¾(y+2)2=4(¾)(x+2)(y+2)2=3(x+2)

Writing an equation of a parabolavertex: (-4, -1)(x + 4)2=4p(y+1)point (2,-4) (2 + 4)2 = 4p(-4 + 1) 36 = -12pp = -3(x + 4)2=4(-3)(y + 1)(x+4)2=-12(y+1)

Writing an equation of a parabolavertex: (2, 1)(y – 1)2=4p(x – 2)point (-1,4) (4 – 1)2 = 4p(-1 – 2) 9 = -12pp = -3/4(y – 1)2=4(-3/4)(x – 2)(y – 1)2=-3(x – 2)

Application Example 1Let the focus F have coordinates (0, p) and the vertex be at (0,0)

equation: x2 = 4pyBecause the reflector is 3ft across and 1 ft deep, the points (1.5, 1) and (-1.5, 1) must lie on the parabola.

(1.5)2 = 4p(1)2.25 = 4pp = .5625ft = 6.75in

So the microphone should be placed inside the reflector along its axis and 6.75 inches from its vertex.

Application Example 2

Let the roadway be the x-axis. Then the vertex is at (500,25)So the equation is:(x – 500)2 = 4p(y – 25)Points (0, 150) and (1000, 150) also on the parabola.

Application Example 2So the equation is:(0 – 500)2 = 4p(150 – 25)250,000 = 500pp = 500(x – 500)2 = 2000(y – 25)Equation for the shape of the main cable.

Application Example 2To find the length of the support cables, solve equation for y

Support cables are every 100ft, so starting with x = 100, then 200, 300, etc.Cables are 105ft, 70ft, 45ft, 30ft, 25ft, 30ft, 45ft, 70ft, 105ft.

2( 500) 252000x y

2( 500) 252000xy

Homework

Page 641: 51-56, 59-63Quiz: