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LECTURE UNIT 004Parabola The locus of a moving point that its distance from a fix point ( ) is equal to its distance from a fixed line ( ). y x = -a Directrix d1 2a Latus Rectum 4a d2 Where: x A v(h, k) F F = Focus 2a v = Vertex, midpoint of the segment d 1 = d2 e = eccentricity = 1 a a 2a If the vertex (h, k) is at the origin, the equation of the parabola is: y2 = 4ax ( ) If the vertex is a horizontal axis parabola ( ) the equation of the parabola is: (y - k)2 = 4a (x - h) If the vertex (h, k) is at the origin, the equation of the parabola is: x2 = 4ay ( ) If the vertex is a vertical axis parabola ( ) the equation of the parabola is: (x - h)2 = 4a (y - k)Cases: y y y y x=-a x=a F y=a v x x v x x v v F F y=-a F (a) y2 = 4ax (b) y2 = -4ax (c) x2 = 4ay (d) x2 = -4aySketch the graph: 1. (y - 2)2 = 4 (x + 3) 2. x2 = -8 (y + 1) 3. (y + 2)2 = -6 (x - 1) “The only active force that arises out of possession is fear of losing the object of possession!”
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The equation x2 + Dx + Ey + F = 0 is a parabola with vertical axis (horizontal directrix) and the equationy2 + Dx + Ey + F = 0 is a parabola with horizontal axis (vertical directrix). To sketch the graph, reduce theequation to standard form. 4. y2 - 2x = 0 5. x2 + 12y - 2x - 11 = 0 6. 2y2 - 5x + 4y - 7= 0 7. y2 + 4x - 6y + 1 = 0 8. x2 - 2y - x= 0 9. y2 - 7x + 3y - 8 = 0 10. x2 - 8y + 4x - 4= 0 11. (x + 2)2 = - (y - 1) 12. y2 - 4x + 2y - 7 = 0 13. x2 + 2y - x = 0Determine the points of intersection and sketch the graph on the same axis. 14. y = x2 + 3x and y = x + 3 15. y = x2 - 3 and y = x2 + 5 16. y = x2 + 5x - 5 and y = x 17. y2 = 2x and x2 = 2y 18. y = x2 and y = 2x + 3Find the equation of the parabola with given conditions. 19. With directrix y = 2 and a focus at (-2, 4). 20. With vertex at the intersection of 3x - y = 7 and x - 2y = 4 and directrix is x = 4. 21. With vertex at the center of the circle x2 + y2 - 3x - 2y = 0 and passing through (-2, -3) with vertical axis. 22. With vertex at the origin and passing through (1, -3) with horizontal, find the length of the latus rectum. “A house is the character of people who live in it”
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PARABOLA Directrix (x=2.5) Example 3: Sketch the graph of (y + 2)2 = -6 (x - 1) Solution: (y + 2)2 = -6 (x - 1) 4|a| = 6 2a 2|a| = 3 a a F |a| = 1.5 v(1,-2) v(1, -2) parabola opens to the left 2a Example 6: 2y2 - 5x + 4y - 7= 0 Solution: Reducing to Standard Form: [2y2 - 5x + 4y - 7= 0] 1 Multiplying both sides of the equation by 1/2 2 2 y - 5 x +2y - 7 = 0 2 2 Directrix (x=2.42) y2 + 2y = 5 x + 7 2 2 Completing squares; y2 + 2y + 1 = 5 x + 7 + 1 2 2 (y + 1)2 = 5 x + 9 v(-1.8,-1) F 2 2 Factoring x; 9 (y + 1)2 = 5 (x + ) 2 5 4|a| = 2.5 2|a| = 1.25 |a| = 0.625 v (-1.8, -1) parabola opens to the right Example 14: y = x2 + 3x and y = x + 3 Solution: By substitution; X2 + 3x = x + 3 X2 + 2x + 3 = 0 (x + 3)(x - 1) = 0 Points of intersection; P1 (-3, 0) and P2 (1, 4) Graph of the parabola; x2 + 3x + 9 = y + 9 4 4 2 (x + 3 ) = y + 9 2 4 v (-1.5, -2.25) Parabola opens upward “For every joy there is a price to be paid”
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Graph of the line; x y + =1 -3 3 P2(1, 4) (0, 3) P1(-3, 0) v(-1.5,-2.25)Example 17: y2 = 2x and x2 = 2ySolution: x2 = 2y By substitution; (2, 2) 2 x2 ( 2 ) = 2x Points of intersection; P1 (0, 0) and P2 (2, 2) Graph of the parabola; v(0, 0) y2 = 2x v(0, 0) parabola opens to the right 2 x = 2y v(0, 0) parabola opens upward y2 = 2xExample 21: With vertex at the center of the circle x2 + y2 - 3x - 2y = 0 and passing through (-2, -3) with vertical axis.Solution: C(- D ,- E ) 2 2 C( 3 ,1) 2 Note that the vertex is at the center of the circle (x - h)2 = 4a (y - k) 2 (x - 3 ) = 4a (y - 1) 2 Solving for a, using P (-2, -3) 2 (-2 - 3 ) = 4a (-3 - 1) 2 a = - 49 64 Therefore; 2 (x - 3 ) = - 16 (y - 1) 2 49 parabola opens downward “The best and shortest road towards knowledge of truth is nature”
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