Engineering Calculus Notes 136

Engineering Calculus Notes 136 - 124 CHAPTER 2. CURVES To...

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Unformatted text preview: 124 CHAPTER 2. CURVES To fix ideas, let us place the y -axis along the directrix and the focus at a point F (k, 0) on the x-axis. The distance of a generic point P (x, y ) from the y -axis is |x|, while its distance from the focus is |F P | = (x − k)2 + y 2 . Thus the focus-directrix property can be written |F P | =e |x| where e is the eccentricity. Multiplying through by |x| and squaring both sides leads to the equation of degree two (x − k)2 + y 2 = e2 x2 . which can be rewritten (1 − e2 )x2 − 2kx + y 2 = −k2 . (2.8) Parabolas When e = 1, the x2 -term drops out, and we have y 2 = 2kx − k2 = 2k x − k 2 . (2.9) Now, we change coordinates, moving the y -axis k/2 units to the right. The effect of this on the equation is a bit counter-intuitive. To understand this, let us fix a point P whose coordinates before the move were (x, y ); let us for a moment denote its coordinates after the move by (X, Y ). Clearly, since nothing moves up or down, Y = y . However, the new origin is k/2 units to the right of the old one—or looking at it another way, the new origin was, in the old coordinate system, at (x, y ) = (k/2, 0), and is now (in the new coordinate system) at (X, Y ) = (0, 0); in particular, X = x − k/2. But this effect applies to all points. So if a point is on our parabola—that is, if its old coordinates satisfy Equation (2.9), then rewriting this in terms of the new coordinates we get Y 2 = 2kX . Switching back to using lower-case x and y for our coordinates after the move, and setting p = 2k, we recover Equation (2.4) y 2 = px ...
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