Stitz-Zeager_College_Algebra_e-book

6 here x h is x 2 so h 2 and y k is y so k 0 hence

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Unformatted text preview: ce from (x, −p) to (x, y ). Using the Distance Formula, Equation 1.1, we get 1 We’ll talk more about what ‘directed’ means later. 408 Hooked on Conics (x − 0)2 + (y − p)2 x2 + (y − p)2 x2 + (y − p)2 x2 + y 2 − 2py + p2 x2 = (x − x)2 + (y − (−p))2 = (y + p)2 = (y + p)2 square both sides 2 + 2py + p2 =y expand quantities = 4py gather like terms 2 Solving for y yields y = xp , which is a quadratic function of the form found in Equation 2.4 with 4 a = 41p and vertex (0, 0). We know from previous experience that if the coefficient of x2 is negative, the parabola opens 2 downwards. In the equation y = xp this happens when p < 0. In our formulation, we say that p is 4 a ‘directed distance’ from the vertex to the focus: if p > 0, the focus is above the vertex; if p < 0, the focus is below the vertex. The focal length of a parabola is |p|. What if we choose to place the vertex at an arbitrary point (h, k )? We can either use transformations (vertical and horizontal shifts from Section 1.8) or re-derive the equation from Definition 7.3 to arrive at the following...
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