**Unformatted text preview: **ly, we verify that one focus is at (0, 0), and the formula given in Exercise 7 in Section
7.5 gives the eccentricity is e in this case as well. If e = 1, the equation r = 1−eed θ) reduces to
cos( r= d
1−cos(θ) which gives the rectangular equation y 2 = 2d x + d
2 . This is a parabola with vertex −d, 0
2 opening to the right. In the language of Section 7.3, 4p = 2d so p = d , the focus is (0, 0),
2
the focal diameter is 2d and the directrix is x = −d, as required. Hence, we have shown that in all
cases, our ‘new’ understanding of ‘conic section’, ‘focus’, ‘eccentricity’ and ‘directrix’ as presented
in Deﬁnition 11.1 correspond with the ‘old’ deﬁnitions given in Chapter 7.
Before we summarize our ﬁndings, we note that in order to arrive at our general equation of a conic
r = 1−eed θ) , we assumed that the directrix was the line x = −d for d > 0. We could have just as
cos(
easily chosen the directrix to be x = d, y = −d or y = d. As the reader can verify, in these cases
we obtain the forms r = 1+eed θ) ,...

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