Transcript Document

LC.02.4 - The General
Equation of Conic
Sections
MCR3U - Santowski
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(A) Review
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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
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(B) Changing Forms of the
Conic Section Equations
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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
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(B) Changing Forms of the
Conic Section Equations
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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
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(C) The General Equation
of Conic Sections
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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
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(D) Example
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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
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(D) Example
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(E) Intersection of Lines
and Conics
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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:
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(E) Intersection of Lines
and Conics
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(F) Example
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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
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(F) Example - Graph
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(G) Homework
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Handout from Nelson text
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