Page 110 - Calc for the Clueless

Find the equation of the parabola with focus (1,3), directrix x = 11.

Drawing F and the directrix, the picture must be the following picture. The vertex is halfway between the x
numbers. So x = (11 + 1)/2 = 6.

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V(6,3). c = the distance between V and F = 5. The equation is (y - 3) = -4c(x - 6) = -20(x - 6>. Remember, the
minus sign is from the shape and c is always positive for these problems.

Example 25—

Vertices (2,3) and (12,3) and one focus (11,3). Find the equation of the ellipse.

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Two vertices give the center, ((12 = 2)/2,3) = (7,3). F(11,3). (x - 7) /a = (y -3) /b = 1. a=12-7-5. c = 11 - 7 = 4.
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a - b2 = c . 5 -b =4 . b =9 (no need for b). (x-7) /25 + (y-3) /9=1.

Example 26—

Find the equation of the hyperbola with vertices (0,±6) and asymptotes y=±(3/2)x.

V(0,±6) says the center is (0,0) and its shape is y /36 - x /a = 1 The slope of the asymptotes is 3/2 =square root
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of the number under y over the square root of the number under the x term. So 3/2 = 6/?. So ? = 4. So a = 4 2
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= 16. The equation is y /36 - x /16 = 1.
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