Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Use Cartesian coordinates to express the equations for the hyperbolas determined by the conditions specified in Exercises 38–43.
$foci (0, \pm 4), directrices y = \pm 2$

The equation is $\frac{y^{2}}{8} - \frac{x^{2}}{8} = 1$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

The given values are,

$foci (0, \pm 4), directrices y = \pm 2$

Step 2. Value of the variables.

Now,

$The focus points are (0, 4), (0, - 4) . Center = (\frac{0 + 0}{2}, \frac{4 - 4}{2}) [since mid point = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})] = (0, 0) Given driectries are y = \pm 2 . That means \frac{b}{e} = 2 e = \frac{b}{2} Then b e = 4 b \cdot \frac{b}{2} = 4 [since e = \frac{b}{2}] b^{2} = 8 c = \sqrt{(0 - 4)^{2} + (0 - 0)^{2}} [since D = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}] c = 4 For a hyperbola, a^{2} + b^{2} = c^{2} a^{2} + 8 = 4^{2} a^{2} = 8$

Step 3. Substitution.

On substituting the obtained values,

$\frac{(y - k)^{2}}{b^{2}} - \frac{(x - h)^{2}}{a^{2}} = 1 where (h, k) is the center. \frac{(y - 0)^{2}}{8} - \frac{(x - 0)^{2}}{8} = 1 [since a^{2} = 8, b^{2} = 8] \frac{y^{2}}{8} - \frac{x^{2}}{8} = 1$

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Calculus

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Short Answer

Use Cartesian coordinates to express the equations for the hyperbolas determined by the conditions specified in Exercises 38–43.
$foci (0, \pm 4), directrices y = \pm 2$

Step by Step Solution

Step 1. Given information.

Step 2. Value of the variables.

Step 3. Substitution.

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Calculus

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Short Answer

Use Cartesian coordinates to express the equations for the hyperbolas determined by the conditions specified in Exercises 38–43. foci (0,±4), directrices y=±2

Step by Step Solution

Step 1. Given information.

Step 2. Value of the variables.

Step 3. Substitution.

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Use Cartesian coordinates to express the equations for the hyperbolas determined by the conditions specified in Exercises 38–43.
$foci (0, \pm 4), directrices y = \pm 2$