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Q 30.

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Precalculus Enhanced with Graphing Utilities
Found in: Page 643
Precalculus Enhanced with Graphing Utilities

Precalculus Enhanced with Graphing Utilities

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.

Vertex is at 4,-2 and focus is at 6,-2.

The equation of a parabola is

y+22=8x-4.

The points are 6,2;6,-6.

The graph of an equation is

See the step by step solution

Step by Step Solution

Step 1. Given Information.

The given vertex is at 4,-2 and focus is at 6,-2.

Step 2. Equation of a parabola.

The vertex is at 4,-2 and focus 6,-2 both lie on the horizontal line y=-2 (the axis of symmetry). The distance a from the vertex to the focus is a=-2.

The parabola opens to the right. The form of a equation is

y-k2=-4ax-h.

where h,k=4,-2 and a=-2. Therefore, the equation is

y+22=8x-4.

Step 3. Latus rectum.

The two points that determines the latus rectum by letting x=6, so that

y+22=8x-4y+22=82y+22=16y+2=±4y+2=4,y+2=-4y=2,-6.

The points are 6,2 and 6,-6.

Step 4. Graphing Utility.

The graph of a parabola is

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