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LC.02.4 - The General Equation of Conic Sections

LC.02.4 - The General Equation of Conic Sections. MCR3U - Santowski. (A) Review. Recall the completing the square technique: If y = 2x 2 – 12x + 5 y = 2(x 2 – 6x + 9 – 9) + 5 y = 2(x – 3) 2 – 18 + 5 So y = 2(x – 3) 2 - 13. (B) Changing Forms of the Conic Section Equations.

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LC.02.4 - The General Equation of Conic Sections

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  1. LC.02.4 - The General Equation of Conic Sections MCR3U - Santowski

  2. (A) Review • Recall the completing the square technique: • If y = 2x2 – 12x + 5 • y = 2(x2 – 6x + 9 – 9) + 5 • y = 2(x – 3)2 – 18 + 5 • So y = 2(x – 3)2 - 13

  3. (B) Changing Forms of the Conic Section Equations • Given the standard equation (x – 2)2/25 – (y + 1)2/36 = -1 (a hyperbola), let’s rewrite this equation in an alternate form by eliminating the fractions and expanding: • To eliminate the denominator, we multiply the equation (i.e. every term in the equation) by 36x25 or 900 • 36(x-2)2 – 25(y+1)2 = -900 • 36(x2 – 4x + 4) – 25(y2+2y+1) + 900 = 0 • 36x2 – 144x + 144 – 25y2 – 50y – 25 + 900 = 0 • 36x2 – 25y2 – 144x – 50y +1019 = 0 • Which is now the general form of the equation for the hyperbola

  4. (B) Changing Forms of the Conic Section Equations • Now given the equation in general form, we can change it back into standard form: • Given 25x2 + 9y2 + 50x – 36y – 164 = 0 • So 25x2 + 50x + 9y2 – 36y = 164 • 25(x2 + 2x + 1 – 1) + 9(y2 – 4y + 4 – 4) = 164 • 25(x + 1)2 + 9(y – 2)2 – 25 – 36 = 164 • 25(x + 1)2 + 9(y – 2)2 = 225 • (x+1)2/9 + (y – 2)2/25 = 1 • And so we have the general equation of our ellipse changed in the standard form of the equation

  5. (C) The General Equation of Conic Sections • For our 4 conic sections (circles, ellipses. Hyperbolas and parabolas), we have the following general equation: • Ax2 + By2 + 2Gx + 2Fy + C = 0 • The values of A and B determine the type of conic section: • If A = B, then we have a circle • If AB > 0 (i.e. both are positive)  ellipse • If AB < 0 (i.e. either are negative)  hyperbola • If AB = 0 (i.e. either are zero)  parabola

  6. (D) Example • Given the conic x2 – y2 + 8x + 4y + 24 = 0, identify and analyze and graph • Firstly, A = 1, B = -1, so AB = (1)(-1) = -1, so we have a hyperbola • (x2 + 8x + 16 – 16) – (y2 + 4y + 4 – 4) = -24 • (x+4)2 – (y+2)2 -20 = -24 • (x+4)2 – (y+2)2 = -4 • (x+4)2/4 – (y+2)2/4 = -1 • Thus we have a hyperbola, centered at (-4,-2) with a = 2 and b = 2 and thus c = 8 = 2.8 • The asymptotes are at y = + 1(x + 4) - 2 (a/b = 2/2 = 1) • The hyperbola opens U/D • The graph follows on the next slide

  7. (D) Example

  8. (E) Intersection of Lines and Conics • Lines and conics can “intersect” in one of three ways  intersect once, not intersect at all, or intersect at two points • Consider the following graphs:

  9. (E) Intersection of Lines and Conics

  10. (F) Example • Given the conic 9x2 + 4y2 - 36 = 0 and the line y = x – 2, find the intersection point(s) • We will use the substitution method for solving this linear/conic system • 9x2 + 4(x – 2)2 – 36 = 0 • 9x2 + 4x2 – 16x + 16 – 20 = 0 • 13x2 – 16x – 20 = 0 • And then we can use the quadratic formula and find that the values for x are 2 and -0.77 • Then 9(2)2 + 4y2 – 36 = 0 so y = 0 • And 9(-0.77)2 + 4y2 – 36 = 0 so y = +2.77

  11. (F) Example - Graph

  12. (G) Homework • Handout from Nelson text

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