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Q. 32

Expert-verified

Calculus

Found in: Page 1

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

In Exercises 24-34 sketch the parametric curve by eliminating the parameter.
$x = \sec t, y = \tan t, t \in (- \frac{π}{2}, \frac{π}{2})$

The graphical representation by using the points $(- 1, 0) (1, 0) (2, - \sqrt{3}) (2, \sqrt{3})$ is as follows,

Therefore, the equation after elimination of the parameter is $x^{2} - y^{2} = 1$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

$x = \sec t, y = \tan t, t \in (- \frac{π}{2}, \frac{π}{2})$

Step 2: Calculation

Consider the parametric equations $x = \sec t, y = \tan t, t \in (- \frac{π}{2}, \frac{π}{2}) .$

The objective is to sketch the parametric curve by eliminating the parameter.

Take the equation $x = \sec t .$

Square the equation on both sides.

$x^{2} = \sec^{2} t$

Take the equation $y = \tan t$ .

Square the equation on both sides. Then.

$y^{2} = \tan^{2} t$

Now subtract the equation $y^{2} = \tan^{2} t$ from $x^{2} = \sec^{2} t$ .

Thus.

$x^{2} - y^{2} = \sec^{2} t - \tan^{2} t x^{2} - y^{2} = 1 [since \sec^{2} t - \tan^{2} t = 1]$

In order to draw the graph of the equation assume $x = - 1, 1, 2$ .

Substitute $x = - 1$ in the equation $x^{2} - y^{2} = 1$ .

Then,

$1^{2} - y^{2} = 1 1 - y^{2} = 1 y = 0 simplify (x, y) = (- 1, 0)$

Substitute $x = 1$ in the equation $x^{2} - y^{2} = 1$

Then.

$(- 1)^{2} - y^{2} = 1 1 - y^{2} = 1 y = 0 simplify (x, y) = (- 1, 0)$

Substitute $x = 2$ in the equation $x^{2} - y^{2} = 1$ .

Then,

$2^{2} - y^{2} = 1 4 - y^{2} = 1$

Add $- 4$ on both sides of the equation.

$4 - y^{2} - 4 = 1 - 4 4 - y^{2} - 4 - - 3 - y^{2} = - 3 y = \sqrt{3} (x, y) = (2, - \sqrt{3}) (2, \sqrt{3})$

The graphical representation by using the points $(- 1, 0) (1, 0) (2, - \sqrt{3}) (2, \sqrt{3})$ is as follows,

Therefore, the equation after elimination of the parameter is $x^{2} - y^{2} = 1$

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