assignment 13

course mth 164

i dint record the numbers given for the last equation and therefore could not complete the last equation.

׫l͛xݷ¥Iassignment #013

013.

Precalculus II

12-17-2008

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13:33:14

Query problem 9.4.12 graph of hyperbola with vertices at (-4,0) and (4,0), asymptote y = 2x

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RESPONSE -->

ok

confidence assessment:

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13:45:02

Give the equation of the specified hyperbola.

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RESPONSE -->

y=0+-2(x-0)

confidence assessment:

** The equation of a hyperbola centered at the origin is

x^2 / a^2 - y^2 / b^2 = 1,

where (a, 0) and (-a, 0) are the vertices and y = +-b / a * x are the asymptotes.

In the present case we have a = 4 and y = 2 x is an asymptote.

It follows that b / a = 2 so that b = 2 a = 2 * 4 = 8.

Thus the equation is

x^2 / 4^2 - y^2 / 8^2 = 1 or

x^2 / 16 - y^2 / 64 = 1. **

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13:46:23

Explain how you know whether the equation is x^2 / a^2 - y^2 / b^2 = 1, x^2 / a^2 + y^2 / b^2 = 1 or -x^2 / a^2 + y^2 / b^2 = 1.

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RESPONSE -->

i know it is -x^2/a^2+y^2/b^2=1 because of the x and y values and the assymtotes.

confidence assessment:

** The point (4, 0) wouldn't make sense in an equation of the form -x^2 / a^2 + y^2 / b^2 = 1, since for this point -x^2 / a^2 would be negative and y^2 / b^2 would be positive, making the left-hand side negative while the right-hand side is 1.

For large x and y the value of x^2 / a^2 + y^2 / b^2 would be large and couldn't be equal to 1.

The form x^2 / a^2 - y^2 / b^2 = 1 makes sense for (4, 0) and also for large x and y, since the difference x^2 / a^2 - y^2 / b^2 could indeed equal 1 even for large x and y. **

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13:46:41

Explain how you obtained the values of a and b for the equation.

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RESPONSE -->

by using the formulas and the assymtotes.

confidence assessment:

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13:51:12

Query problem 9.4.24 graph of hyperbola defined by rectangle of width 4, centered at the origin, lines y = + - 2x passing through diagonals, opening right and left.

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RESPONSE -->

ok

confidence assessment:

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13:54:07

Give the equation of the specified hyperbola.

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RESPONSE -->

the equation was

(x-h)^2- (y-k)^2

A^2 b^2 =1

but i dont remember the preceeding numbers to plug in

** The width of the rectangle is 4 so its vertical sides are at x = -2 and x = 2.

Since the parabola opens to the right the vertices are on the right and left sides of the rectangle, at (-2, 0) and (2, 0).

The lines y = +- 2 x are the asymptotes.

The form of the equation is x^2 / a^2 - y^2 / b^2 = 1, with vertices (a, 0) and (-a, 0) and asymptotes y = +- b / a * x.

It follows that a = 2 (vertices at (-a,0) and (a,0)) and b / a = 2 (since y = +-2 x = +- b / a * x).

We get b = 2 a = 2 * 2 = 4 so the equation is

x^2 / 2^2 - y^2 / 4^2 = 1 or

x^2 / 4 - y^2 / 16 = 1. **

confidence assessment:

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13:58:11

Explain how you know whether the equation is x^2 / a^2 - y^2 / b^2 = 1, x^2 / a^2 + y^2 / b^2 = 1 or -x^2 / a^2 + y^2 / b^2 = 1.

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RESPONSE -->

the equation is one of the following because i know this is the correct order but as stated earlier i dont remember the figures to correctly answer

confidence assessment:

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13:58:43

Explain how you obtained the values of a and b for the equation.

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RESPONSE -->

you use the x coordinates to find your a and b

confidence assessment:

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assignment #013

013.

Precalculus II

12-17-2008"