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Today in Precalculus. Go over homework Notes: Parabolas Completing the square Writing the equation given the graph Applications Homework. Example 1. Prove that the graph of y 2 + 2y – 8x – 7 = 0 is a parabola y 2 + 2y = 8x + 7 y 2 + 2y + 1 = 8x + 7 + 1 y 2 + 2y + 1 = 8x + 8

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Today in Precalculus

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Today in precalculus

Today in Precalculus

  • Go over homework

  • Notes: Parabolas

    • Completing the square

    • Writing the equation given the graph

    • Applications

  • Homework


Example 1

Example 1

Prove that the graph of y2 + 2y – 8x – 7 = 0 is a parabola

y2 + 2y = 8x + 7

y2 + 2y + 1 = 8x + 7 + 1

y2 + 2y + 1 = 8x + 8

(y+1)2 = 8 (x + 1)

Standard form for a parabola


Example 11

Example 1

(y+1)2 = 8 (x + 1)

Graph:

Opens to the right

vertex (-1, -1)

4p = 8, p = 2

focal length =2

focal width = 8

focus: (1, -1)

directrix: x = -3

axis: y = -1


Example 2

Example 2

Prove that the graph of x2 – 2x + 3y + 7 = 0 is a parabola

x2 – 2x = -3y – 7

x2 – 2x + 1 = -3y – 7 + 1

x2 – 2x + 1 = -3y – 6

(x – 1)2 = -3(y + 2)

Standard form for a parabola


Example 21

Example 2

(x – 1)2 = -3(y + 2)

Graph:

Opens downward

vertex (1, - 2)

p = -3/4

focal length = -3/4

focal width = 3

focus: (1, -11/4)

directrix: y = -5/4

axis: x = 1


Writing an equation of a parabola

Writing an equation of a parabola

Find 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 p

9 = 12p

p = ¾

(y+2)2=4(¾)(x+2)

(y+2)2=3(x+2)


Writing an equation of a parabola1

Writing an equation of a parabola

vertex: (-4, -1)

(x + 4)2=4p(y+1)

point (2,-4)

(2 + 4)2 = 4p(-4 + 1)

36 = -12p

p = -3

(x + 4)2=4(-3)(y + 1)

(x+4)2=-12(y+1)


Writing an equation of a parabola2

Writing an equation of a parabola

vertex: (2, 1)

(y – 1)2=4p(x – 2)

point (-1,4)

(4 – 1)2 = 4p(-1 – 2)

9 = -12p

p = -3/4

(y – 1)2=4(-3/4)(x – 2)

(y – 1)2=-3(x – 2)


Application example 1

Application Example 1

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

equation: x2 = 4py

Because 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 = 4p

p = .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

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 21

Application Example 2

So the equation is:

(0 – 500)2 = 4p(150 – 25)

250,000 = 500p

p = 500

(x – 500)2 = 2000(y – 25)

Equation for the shape of the main cable.


Application example 22

Application Example 2

To 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.


Homework

Homework

Page 641: 51-56, 59-63

Quiz:


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