James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 10.5 Conic Sections 675
y
F(0, p)
O
P(x, y)
y=_p
y
p
x
then the distance from P to the focus is
| PF | − sx 2 1 sy 2 pd 2
and the distance from P to the directrix is | y 1 p | . (Figure 3 illustrates the case where
p . 0.) The defining property of a parabola is that these distances are equal:
sx 2 1 sy 2 pd 2 − | y 1 p |
We get an equivalent equation by squaring and simplifying:
FIGURE 3
x 2 1 sy 2 pd 2 − | y 1 p | 2 − sy 1 pd 2
x 2 1 y 2 2 2py 1 p 2 − y 2 1 2py 1 p 2
x 2 − 4py
1 An equation of the parabola with focus s0, pd and directrix y − 2p is
x 2 − 4py
If we write a − 1ys4pd, then the standard equation of a parabola (1) becomes y − ax 2 .
It opens upward if p . 0 and downward if p , 0 [see Figure 4, parts (a) and (b)]. The
graph is symmetric with respect to the y-axis because (1) is unchanged when x is replaced
by 2x.
y
y
y
y
(0, p)
0 x
y=_p
0
(0, p)
y=_p
x
x=_p
( p, 0)
0 x
( p, 0)
0 x
x=_p
FIGURE 4
(a) ≈=4py, p>0
(b) ≈=4py, p<0
(c) ¥=4px, p>0
If we interchange x and y in (1), we obtain
(d) ¥=4px, p<0
2
y 2 − 4px
y
¥+10x=0
5
”_ , 0’ 2
FIGURE 5
0 x
x= 5 2
which is an equation of the parabola with focus sp, 0d and directrix x − 2p. (Interchanging
x and y amounts to reflecting about the diagonal line y − x.) The parabola
opens to the right if p . 0 and to the left if p , 0 [see Figure 4, parts (c) and (d)]. In both
cases the graph is symmetric with respect to the x-axis, which is the axis of the parabola.
Example 1 Find the focus and directrix of the parabola y 2 1 10x − 0 and sketch
the graph.
SOLUtion If we write the equation as y 2 − 210x and compare it with Equation 2, we
see that 4p − 210, so p − 2 5 2 . Thus the focus is s p, 0d − s2 5 2 , 0d and the directrix is
x − 5 2 . The sketch is shown in Figure 5.
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