Stitz-Zeager_College_Algebra_e-book

Solution we need only complete the square on the x

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Unformatted text preview: meter of the parabola is |4p|, which, in light of Equation 7.2, is easy to find. In our last example, for instance, when graphing (x + 1)2 = −8(y − 3), we can use the fact that the focal diameter is | − 8| = 8, which means the parabola is 8 units wide at the focus, to help generate a more accurate graph by plotting points 4 units to the left and right of the focus. Example 7.3.2. Find the standard form of the parabola with focus (2, 1) and directrix y = −4. Solution. Sketching the data yields, 2 3 No, I’m not making this up. Consider this an exercise to show what follows. 410 Hooked on Conics y 1 x −1 −1 1 2 3 The vertex lies on this vertical line midway between the focus and the directrix −2 −3 From the diagram, we see the parabola opens upwards. (Take a moment to think about it if you don’t see that immediately.) Hence, the vertex lies below the focus and has an x-coordinate of 2. To find the y -coordinate, we note that the distance from the focus to the directrix is 1 − (−4) = 5, which means the vertex lie...
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