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Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
Derive the equation x² = 4py in Figure 10.4.6.

Transcribed Image Text:PARABOLAS IN STANDARD POSITION
y =p
(0, p)
(0, -p)
(p. 0)
(-p. 0)
X =-P
= 4px
–4px
x* = 4py
x² = -4py
Figure 10.4.6
To illustrate how the equations in Figure 10.4.6 are obtained, we will derive the equation
for the parabola with focus (p, 0) and directrix x = -p. Let P(x, y) be any point on the
parabola. Since P is equidistant from the focus and directrix, the distances PF and PD in
Figure 10.4.7 are equal; that is,
PF = PD
(1)
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