686 CHAPTER 10 Analytic Geometry Figure 11 =− y x8 2 x y V (22, 4) (22, 24) (0, 0) 5 25 5 25 1 – 2 (2 , 2) F 5 (22, 0) D: x 5 2 Because the point ( ) − 1 2 , 2 is on the parabola, the coordinates = − = x y 1 2 , 2 satisfy = − y ax 4 . 2 Substituting = − x 1 2 and = y 2 into the equation leads to ( ) = − − = a a 2 4 1 2 2 2 The equation of the parabola is = − ⋅ = − y x x 4 2 8 2 The focus is ( ) −2, 0 and the directrix is the line = x 2. Substituting = − x 2 in the equation = − y x8 2 gives = ± y 4. The points ( ) −2, 4 and ( ) − − 2, 4 determine the latus rectum. See Figure 11. Now Work PROBLEM 31 2 Analyze Parabolas with Vertex at h k , ( ) If a parabola with vertex at the origin and axis of symmetry along a coordinate axis is shifted horizontally h units and then vertically k units, the result is a parabola with vertex at ( ) h k , and axis of symmetry parallel to a coordinate axis. The equations of such parabolas have the same forms as those in Table 1, but with x replaced by −x h (the horizontal shift) and y replaced by −y k (the vertical shift). Table 2 gives the forms of the equations of such parabolas. Figures 12(a)–(d) illustrate the graphs for > > h k 0, 0. TIP Rather than memorizing Table 2, use transformations (shift horizontally h units, vertically k units) and the fact that a is the distance from the vertex to the focus to determine the parabola. j Equations of a Parabola: Vertex at , ; h k ( ) Axis of Symmetry Parallel to a Coordinate Axis; 0 a > Vertex Focus Directrix Equation Description h k , ( ) h k , ( ) + a x h = −a y k x h 4 2 ( ) ( ) − = − a Axis of symmetry is parallel to the x-axis, the parabola opens right h k , ( ) h k , ( ) −a x h = + a y k x h 4 2 ( ) ( ) − = − − a Axis of symmetry is parallel to the x-axis, the parabola opens left h k , ( ) h k , ( ) + a y k = −a x h y k 4 2 ( ) ( ) − = − a Axis of symmetry is parallel to the y-axis, the parabola opens up (is concave up) h k , ( ) h k , ( ) −a y k = + a x h y k 4 2 ( ) ( ) − = − − a Axis of symmetry is parallel to the y-axis, the parabola opens down (is concave down) Table 2 Figure 12 y V 5 (h, k) Axis of symmetry x 5 h x y V 5 (h, k) Axis of symmetry x 5 h y V 5 (h, k) Axis of symmetry y 5 k x y V 5 (h, k) Axis of symmetry y 5 k F 5 (h, k 1 a) D: y 5 k 2 a F 5 (h, k 2 a) D: y 5 k 1 a F 5 (h 2 a, k) D: x 5 h 1 a F 5 (h 1 a, k) D: x 5 h 2 a (c) (x 2 h) 2 5 4a(y 2 k) (d) (x 2 h) 2 5 24a(y 2 k) (b) (y 2 k) 2 5 24a(x 2 h) (a) (y 2 k) 2 5 4a(x 2 h) x x Finding the Equation of a Parabola, Vertex Not at the Origin Find an equation of the parabola with vertex at ( ) −2, 3 and focus at ( ) 0, 3 . Graph the equation. EXAMPLE 7 x y 1 2 , 2 =− = Solve for a.
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