You can see from the diagram why we call these shapes conic sections.
Conic sections -- usually just called conics -- are the four types of curve you see above
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An ellipse is the locus or path of all points such that the SUM of the distances from 2 fixed points called the foci is constant and equal to the measure of the major axis.
The equation of the ellipse center (0, 0) with semi-major axis = a and semi-minor axis = b is:
If we move the center to (h, k), the equation becomes:
Definition: the chord through a focus perpendicular to the major axis is called the latus rectum.
Important Note Though we've labeled the semi-major axis "a" throughout history, the sages who wrote the Quebec textbooks decided that "a" will always be the horizontal distance from the center to a vertex. Keep this in mind when doing vertical ellipses. Because of this wisdom, we need to learn 2 different formulas for c2 , since a is sometimes less than b. To get around this, just use the difference between the larger and the smaller for c 2.
Example 1
Find the equation of the ellipse with center (-2, 3) through vertices (-6, 3) and (-2, 5).
Solution
we can see that a = 4 (since (-6, 3) is 4 units from center) and b = 2.
The equation of this ellipse is:
Now let's do a vertical ellipse.
Example 2
Find the equation of the ellipse with center (4, 6), major axis = 4, parallel to the y-axis, and semi-minor axis = 1. List the coordinates of the foci and vertices.
Solution b = 1 and a = 2 so, c 2 = b 2 - a 2 so, c = 
the equation is:
The foci are at (4, 
), the vertices are at (4, 8) (4, 4) (3, 6) and (5, 6).
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A hyperbola is the locus or path of all points such that the absolute value of the difference of the distances from 2 fixed points called the foci is a constant.
The equation of a hyperbola is identical to that of an ellipse except there is a minus sign instead of a plus sign.
Note Just as with the ellipse, there are 2 axes in a hyperbola. They are called the transverse and conjugate axes. The transverse axis crosses the hyperbola, (through the vertices), and the conjugate axis is perpendicular to the transverse axis.
The dotted lines through the center of the hyperbola are the asymptotes.
The equation of the hyperbola with center (h, k), with transverse axis = 2a and
conjugate axis = 2b is: 
Note
If the hyperbola's transverse axis is horizontal, the equation = + 1.
If the hyperbola's transverse axis is vertical, the equation = -1.
The equations of the asymptotes to the hyperbola center (h, k) are:
(y - k) = 
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Example 3
Find the equation of the hyperbola with transverse axis = 12, parallel to the y-axis, conjugate axis = 6, center at (3, -8). Find the coordinates of the foci, the vertices, and the equations of the asymptotes.
Since the transverse axis is parallel to the y-axis, the hyperbola is vertical so we'll use -1.
Since the transverse axis = 12, b = 6
Since the conjugate axis = 6, a = 3
The equation of this hyperbola is: 
Since the center is (3, -8), the vertices will be (0, -8), (6,-8), (3, -2) and (3, -14).
The foci will be on the line x = 3 and we'll find c to get their coordinates.
c2 = a2 + b2 so c = 
so, the coordinates of the foci are (3, -8 
).
The equations of the asymptotes are: (y + 8) = 
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Equations from General form to Standard Form
To change the equation of an ellipse or hyperbola from general to standard form, we must complete the squares on both the x and y terms.
Example Put this ellipse equation in standard form, find the coordinates of the center, foci, and vertices.
4x2 + y2 - 8x - 2y + 1 = 0 u 4x2 - 8x + y2 - 2y = -1
4(x2 - 2x) + y2 - 2y = -1 u 4(x - 1)2 + (y - 1)2 = -1 + 4 + 1
4(x - 1)2 + (y - 1)2 = 4 u
center (1, 1), c = 
so, foci are at (1, 1
)
vertices at (0, 1) , (2, 1), (1, 3), and (1, -1).
Example 4
Put this hyperbola equation in standard form. Find the coordinates of the center and foci, and the equations of the asymptotes.
x2 - y 2 - 2x - 4y - 4 = 0 
x2 - 2x - y 2 - 4y = 4
(x - 1)2 - (y + 2)2 = 4 + 1 - 4 
(x - 1)2 - (y + 2)2 = 1
(this is like a circle that was cut in half and the sides turned back to back.)
center (1, -2), c = 
so, foci (1
, -2), vertices are (2, -2) and (0, -2).
the asymptotes are y = 
x.
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Though you covered the parabola in 436, you never learned the definition nor the features of the parabola. We'll do that now.
A parabola is the set of all points that are
equidistant from a fixed point called the focus and
a fixed straight line called the directrix.
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FOUR PROPERTIES OF THE PARABOLA:
1) The vertex lies midway between the focus and the directrix.
2) The parabola is symmetric about the line VF -- the line through the focus and the vertex, perpendicular to the directrix called the axis of symmetry.
3) A parabola continues infinitely in both directions.
4) The parabola exists on only one side of the directrix -- the side on which the focus lies.
Note In the above diagram, p is the focal length. (some books use c)
Note When the x-term is squared, the parabola opens up or down. When the y-term is squared, the parabola opens left or right.
The parabola vertex (h, k)
When we move the vertex from (0, 0) to (h, k), the equations become:
(x - h) 2 = 
4p(y - k) for parabolas opening up or down.
(y - k) 2 = 
4p(x - h) for parabolas opening right or left.
Example Put this parabola equation in standard form, list the vertex, focus, equation of the directrix and graph.
x 2 + 6x + 4y + 5 = 0
x 2 + 6x = -4y - 5
(x + 3) 2 = -4y - 5 + 9 
(x + 3) 2 = -4(y - 1)
The vertex is at (-3, 1) and p = 1 so, the focus is at (-3, 0) the directrix is y = 2.
Example 5
Put this parabola equation in standard form, list the vertex, focus, equation of the directrix and draw the graph.
y 2 + 6y - 8x - 31 = 0
y 2 + 6y = 8x + 31 
( y + 3) 2 = 8x + 31 + 9
( y + 3) 2 = 8(x + 5)
Since the y-term is squared, the parabola opens to the side. (the right)
vertex (-5, -3) p = 2 focus (-3, -3) directrix: x = -7
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A/ Ellipse
1) Find the equation and the coordinates of the foci for an ellipse with
centre (2, -1) and vertices at (5, -1), (2, 1), (-1, -1) and (2, -3). (make a diagram)
2) An ellipse has major axis of 12 units, and foci at (-4, -1) and (-4, 7). (make a diagram)
a) Find the equation.
b) List the coordinates of the vertices.
c) List the coordinates of the endpoints of the latus rectums
d) find the measure of the latus rectum.
3) Find the centre, vertices and foci of the ellipse whose equation is
x 2 + 4y 2 - 10x + 16y = - 37
4) Write the equation of the largest possible ellipse that can be drawn on a piece of paper measuring 120 cm by 240 cm. (hint: use a diagram with the midpoint of the page as the origin)
B/ Hyperbola
5)
| a) Write the equation in standard form | b) list the center, vertices, and foci, |
| c) describe the curve (see note below) | d) draw the graph |
| 1) 9x 2 - 4y 2 - 54x - 40y - 55 = 0 | 2) 16x 2 - 9y 2 + 64x + 18y + 199 = 0 |
note: to describe the curve, give the equations of axes and asymptotes, state whether the curve opens vertically or horizontally.
6) What is the distance between the foci of the hyperbola
C/ Parabola
7) For the parabolic equations listed below:
| a) find the coordinates of the vertex and focus | b) write the equation of the axis of symmetry |
| c) write the equation of the directrix | d) draw the graph |
| 1) (x - 2) 2 = 8(y - 1) | 2) (y + 4) 2 = - 12 (x + 1) |
ConicMix
8) It is the year 2015 wherein Jaclyn and Norman have opened an automotive design centre. They are working on a futuristic car seen in the diagram.
The body of the car is an ellipse, centre (0, 0), major axis = 12 dm, minor axis = 6 dm.
The wheels are circles, centers at the endpoints of the latus rectum (ellipse),
radius = ½ the latus rectum.
The roof, a parabola, rises 2.5 dm. above the body,
touches the ellipse at the endpoints of the latus rectum.
The bumpers are a hyperbola, centre (0, 0), sharing vertices with the ellipse which forms the body, foci 2 dm. from the vertices.
Find the equations of all the curves and the coordinates of all the vertices, foci and centers of the conics that define the shape of the car.
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1) Find the equation and the coordinates of the foci for an ellipse with
centre (2, -1), vertices at (5, -1), (2, 1), (-1, -1) and (2, -3).
a = 3 (distance from centre to vertex), b = 2
equation is 
c 2 = a 2 + b 2 u c 2 = 5, and c, the focal distance = 
so, the foci are at 
2) An ellipse has major axis of 12 units, and foci at (-4, -1) and (-4, 7).
This says that the ellipse is vertical, the distance between the foci is 8 units, so c = 4
Since the major axis = 12, a = 6, and c 2 = a 2 - b 2 u b 2 = 36 - 16 = 20
a) Find the equation.
The equation is 
b) List the coordinates of the vertices. (as shown in drawing)
and the ends of the horizontal axis are at 
c) List the coordinates of the endpoints of the latus rectums
We find the intersection of y = 7 and y = -1 with the ellipse.
The endpoints of the upper latus rectum are 
The endpoints of the upper latus rectum are
d) find the measure of the latus rectum. l = 20/3 units.
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3) Find the centre, vertices and foci of the ellipse whose equation is
x 2 + 4y 2 - 10x + 16y = - 37
x 2 - 10x + 4y 2 + 16y = - 37
(x - 5) 2 + 4(y 2 + 4y) = -37 + 25
(x - 5) 2 + 4(y + 2)2 = -37 + 25 + 16 = 4
Centre is (5, -2), ellipse is horizontal, vertices at (3, -2), (7, -2), (5, -3), (5, -1)
The foci are at 
4) Write the equation of the largest possible ellipse that can be drawn on a piece of paper measuring 120 cm by 240 cm. (hint: use a diagram with the midpoint of the page as the origin)
The semi-major axis a = ½(240) = 120
The semi-minor axis b = ½(120) = 60
The equation of the ellipse is:
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B/ Hyperbola
5)
| a) Write the equation in standard form | b) list the center, vertices, and foci, | ||
| c) describe the curve (see note below) | d) draw the graph | ||
| 1) 9x 2 - 4y 2 - 54x - 40y - 55 = 0 | 2) 16x 2 - 9y 2 + 64x + 18y + 199 = 0 | ||
| 9(x 2- 6x) - 4(y 2 + 10y) = 55 | 16(x 2+ 4x) - 9(y 2+ 2y) = -199 | ||
| 9(x - 3) 2 - 4(y + 5) 2 = 55 + 81 - 100 = 36 | 16(x + 2) 2 - 9(y + 1) 2 = 64 -9 - 199 = 144 | ||
c 2 = a 2 + b 2 = 13 u c =  |
c 2 = a 2 + b 2 = 25 u c = 5 | ||
1 a) equation is  |
2 a) equation is  | ||
| b) center (3, -5) | b) center (-2, -1) | ||
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 |
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| c) transverse: y = -5, conjugate: x = 3 | c) transverse: x = -2, conjugate: y = -1 | ||
| asymptotes equations:
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horizontal hyperbola | asymptotes equations:
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vertical hyperbola |
| d) you have all the info, draw the picture. | |||
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6) a 2 = 16, and b 2 = 9
for a hyperbola, c 2 = a 2 + b2 so c 2 = 25
the distance between the foci is 2c, therefore 10 units between the foci.
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7) For the parabolic equations listed below:
| a) find the coordinates of the vertex and focus | b) write the equation of the axis of symmetry |
| c) write the equation of the directrix | d) draw the graph |
| 1) (x - 2) 2 = 8(y - 1), opens upwards | 2) (y + 4) 2 = - 12 (x + 1), opens left |
| the form is (x - h) 2 = 4p(y - k), p = 2 | the form is (y - k) 2 = - 4p(x - h), p = 3 |
| a) vertex (h, k) is (2, 1), focus (2, 3) | a) vertex (h, k) is (-1, -4), focus (-4, -4) |
| b) x = 2 is axis of symmetry | b) y = -4 is axis of symmetry |
| c) y = -1 is directrix | c) x = 2 is directrix |
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d) diagrams
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8)
ellipse: a = 6, b = 3 c =  |
circles: center are ( r =  |
 Foci at  |
 |
set x =  |
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parabola: vtx. at (0,  , P(!  |
hyperbola: a = 6, c = 8 so b 2 = 28 |
 |
so  |
  |
Foci ( , asym: y =  |
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