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Q 23.

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Precalculus Enhanced with Graphing Utilities

Found in: Page 643

Book edition 6th

Author(s) Sullivan

Pages 1200 pages

ISBN 9780321795465

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Short Answer

Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation by hand.
Focus at $(- 2, 0)$ and directix of the line $x = 2$ .

The equation of a parabola is $y^{2} = - 8 x$ . The points are $(- 2, 4)$ and $(- 2, - 4)$ . The graph of an equation is :

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information.

The given focus of an equation be $(- 2, 0)$ and an equation of its directix of a line be $x = 2$ .

Step 2. Equation of a parabola.

The vertex of an equation be at $(0, 0)$ . The vertex is midway between the focus and the directix. Since, the focus is on the x-axis at $(- 2, 0)$ , the equation of the parabola is f the form

role="math" localid="1646657517636" $y^{2} = 4 a x$ with $a = - 2$ .

i.e. $y^{2} = - 8 x$ .

Step 3. Latus rectum.

The points that determine the latus rectum by letting $x = - 2$ .

$y^{2} = - 8 x y^{2} = - 8 (- 2) y^{2} = 16 y = \pm 4$ .

The points are $(- 2, 4)$ and $(- 2, - 4)$ .

Step 4. Graphing utility.

The graph of an equation $y^{2} = - 8 x$ will be

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