1 LC The General Equation of Conic Sections MCR3U - Santowski

2 (A) Review Recall the completing the square technique: Recall the completing the square technique: If y = 2x 2 – 12x + 5 If y = 2x 2 – 12x + 5 y = 2(x 2 – 6x + 9 – 9) + 5 y = 2(x 2 – 6x + 9 – 9) + 5 y = 2(x – 3) 2 – y = 2(x – 3) 2 – So y = 2(x – 3) So y = 2(x – 3)

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: 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 To eliminate the denominator, we multiply the equation (i.e. every term in the equation) by 36x25 or (x-2) 2 – 25(y+1) 2 = (x-2) 2 – 25(y+1) 2 = (x 2 – 4x + 4) – 25(y 2 +2y+1) = 0 36(x 2 – 4x + 4) – 25(y 2 +2y+1) = 0 36x 2 – 144x – 25y 2 – 50y – = 0 36x 2 – 144x – 25y 2 – 50y – = 0 36x 2 – 25y 2 – 144x – 50y = 0 36x 2 – 25y 2 – 144x – 50y = 0 Which is now the general form of the equation for the hyperbola 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: Now given the equation in general form, we can change it back into standard form: Given 25x 2 + 9y x – 36y – 164 = 0 Given 25x 2 + 9y x – 36y – 164 = 0 So 25x x + 9y 2 – 36y = 164 So 25x x + 9y 2 – 36y = (x 2 + 2x + 1 – 1) + 9(y 2 – 4y + 4 – 4) = (x 2 + 2x + 1 – 1) + 9(y 2 – 4y + 4 – 4) = (x + 1) 2 + 9(y – 2) 2 – 25 – 36 = (x + 1) 2 + 9(y – 2) 2 – 25 – 36 = (x + 1) 2 + 9(y – 2) 2 = (x + 1) 2 + 9(y – 2) 2 = 225 (x+1) 2 /9 + (y – 2) 2 /25 = 1 (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 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: For our 4 conic sections (circles, ellipses. Hyperbolas and parabolas), we have the following general equation: Ax 2 + By 2 + 2Gx + 2Fy + C = 0 Ax 2 + By 2 + 2Gx + 2Fy + C = 0 The values of A and B determine the type of conic section: The values of A and B determine the type of conic section: If A = B, then we have a circle If A = B, then we have a circle If AB > 0 (i.e. both are positive)  ellipse 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 negative)  hyperbola If AB = 0 (i.e. either are zero)  parabola If AB = 0 (i.e. either are zero)  parabola

6 (D) Example Given the conic x 2 – y 2 + 8x + 4y + 24 = 0, identify and analyze and graph Given the conic x 2 – y 2 + 8x + 4y + 24 = 0, identify and analyze and graph Firstly, A = 1, B = -1, so AB = (1)(-1) = -1, so we have a hyperbola Firstly, A = 1, B = -1, so AB = (1)(-1) = -1, so we have a hyperbola (x 2 + 8x + 16 – 16) – (y 2 + 4y + 4 – 4) = -24 (x 2 + 8x + 16 – 16) – (y 2 + 4y + 4 – 4) = -24 (x+4) 2 – (y+2) = -24 (x+4) 2 – (y+2) = -24 (x+4) 2 – (y+2) 2 = -4 (x+4) 2 – (y+2) 2 = -4 (x+4) 2 /4 – (y+2) 2 /4 = -1 (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 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 asymptotes are at y = + 1(x + 4) - 2 (a/b = 2/2 = 1) The hyperbola opens U/D The hyperbola opens U/D The graph follows on the next slide 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 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: Consider the following graphs:

9 (E) Intersection of Lines and Conics

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

11 (F) Example - Graph

12 (G) Homework Handout from Nelson text Handout from Nelson text