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Section 3.2 - Circles
Conic Sections:
Circles
The equation of a circle with center ),( kh and radius r is
222 )()( rkyhx
Sketch the graph of the circle or semicircle
Important measurements/ points of a circle?
#24 2 2 7x y
Exercise: (π₯ + 4)2 + (π¦ β 2)2 = 25
2
Find an equation of the circle that satisfies the stated conditions
#36 Center 4,1 , radius 5C
#40 Center at the origin, passing through 4, 7P
#46 Find the equation of the circle that satisfies the following conditions:
Endpoints of a diameter π΄ (β5, 2) and π΅ (3, 6)
The technique of completing the square:
Exercise: Solve by completing the square:
ππ + ππ β ππ = π
3
By completing the square for x and y, find the equation of the circle:
#48 2 2 8 10 37 0x y x y
#52 2 29 9 12 6 4 0x y x y
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Section 11.1 β Parabolas (p.734)
Definition of a Parabola
A parabola is the set of all points in a plane equidistant from a fixed point F (the focus) and a fixed line l (the
directrix) that lie in the plane, with F not on l.
If π < 0, the parabola opens downwards
If π < 0, the parabola opens to the left.
The opening width at the focus is 4p,
it is called the latus rectum
5
Find the vertex, focus, and directrix of the following parabolas. Sketch the graph, showing the focus and the
directrix.
#4 220x y
#6 2 1
3 12
x y
#11 π¦2 + 14π¦ + 4π₯ + 45 = 0
6
Find an equation of the parabola that satisfies the given conditions
#24 Focus πΉ(β3, β2), directrix π¦ = 1
#26 Vertex 2,3V , directrix π₯ = 1
#32 Vertex (4, 7) , focus πΉ(4, 2)
#35 Vertex π(β3, 5), axis parallel to the π₯-axis, and passing through the point (5, 9).
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#36 Find an equation of the parabola that has vertex π(3, β2), axis parallel to the π₯-axis, and
π¦-intercept (0,1)
#38 Find an equation for the set of points in an π₯π¦-plane that are equidistant from the point π(7, 0) and
the line with the equation π₯ = 1
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Section 11.2 β Ellipses (p.746)
Definition of an Ellipse
An ellipse is the set of all points in a plane, the sum of whose distances from two fixed
points (the foci) in the plane is a positive constant.
π, π, π > 0 (they are distances, they cannot be negative)
The ratio between c and a is called the eccentricity, the
eccentricity, π =π
π
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Find the major axis, minor axis, vertices, and foci of the following ellipses and graph them
#2 π₯2
25+
π¦2
16= 1
#10 (π₯+2)2
25+
(π¦β3)2
4= 1
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Exercise 1: Find the standard equation of the ellipse with
general equation
16π₯2 + 9π¦2 + 96π₯ β 72π¦ + 144 = 0, then find the
center, vertices and foci of the ellipse.
Exercise 2: Find an equation for the ellipse that satisfies the given conditions: center πΆ(1, β2) with
horizontal axis length of 4 and vertical axis length of 8 .
#20 Find an equation for the ellipse that has its center at the origin and has vertices π(0, Β±7) and foci
πΉ(0, Β±2).
Exercise 2: Find an equation for the ellipse that has foci πΉ(7, β2) and πΉ(1, β2) and a vertex at π(8, β2).
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#25 Find an equation for the ellipse that has its center at the origin, vertices π(0, Β±6) and passes
through the point (3, 2).
#28 Find an equation for the ellipse that has its center at the origin, passing through (2, 8) and (4, 4).
#30 Find an equation for the ellipse that has its center at the origin with eccentricity 4
7 and vertices
π(Β±7, 0)
#42 Find an equation for the set of points in an π₯π¦-plane such that the sum of the distances from
πΉ(12, 0) and πΉβ²(β12, 0) is π = 26 when
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Section 11.3 β Hyperbolas (p.758)
Definition of a hyperbola A hyperbola is the set of all points in a plane, the difference of
whose distances from two fixed points (the foci) in the plane is a
positive constant.
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Find the vertices, the foci, and the equation of the asymptotes of the hyperbola. Sketch its graph, showing
the asymptotes and the foci.
#2 π¦2
49β
π₯2
16= 1
Exrcise 1: π₯2
16β
π¦2
49= 1
#12 (π₯β3)2
25β
(π¦β1)2
4= 1
#16 Find the standard equation as well as the vertices, the foci, and the equation of the asymptotes of
the hyperbola. Sketch its graph, showing the asymptotes and the foci.
25π₯2 β 9π¦2 + 100π₯ β 54π¦ + 10 = 0
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#22 Find an equation for the hyperbola that has its center at the origin and foci πΉ(Β±8, 0) and vertices
π(Β±5, 0).
#31 Foci 0, 10F , asymptotes 1
3y x
Exercise 2: Find an equation for the hyperbola that has vertices π(β2, 5) and π(β2, 1) as well as a
focus πΉ(β2, 6).
Exercise 3: Identify the graphs of the following equations as parabola, circle, ellipse or hyperbola. Identify
the orientation (horizontal or vertical)
a. 5π₯2 β 7π¦2 + 30π₯ + 28π¦ β 18 = 0
b. 3π₯2 + 3π¦2 β 12π₯ + 6π¦ β 60 = 0
c. π₯2 + 6π₯ β 8π¦ + 25 = 0
d. 4π₯2 β 4π¦2 β 24π₯ β 16π¦ β 52 = 0
e. 5π₯2 + 7π¦2 + 40π₯β 28π¦ + 3 = 0
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Section 11.5 β Polar Coordinates (p.784)
Instead of expressing points in the coordinate plane by distance in x- and y- direction, as ,P x y , points
are expressed by the radius of a circle with center at the origin O (the pole) and the angle between the
positive x-axis and the line OP. As usual, is positive, if it is measured in counterclockwise direction from
the x-axis. The point is now expressed as ,P r .
The origin or pole has the coordinates 0,O for any .
Polar coordinates are not unique!
Graph the points π1(4,3π
4) and π2 (β4,
7π
4)
in the given polar coordinate system:
βΉ polar coordinates can have a negative radius.
This means, we start measuring the angle from
the negative π₯ β axis
#1 Which polar coordinates represent the same point as (3,π
3)?
(a) (3,7π
3) (b) (3, β
π
3)
(c) (β3,4π
3) (d) (3, β
2π
3)
(e) (β3, β2π
3) (f) (β3, β
π
3)
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Relationship between rectangular and polar coordinates:
If π(π₯, π¦) and π(π, π) determine the same point π, then
(1) π₯ = π cos π , π¦ = π sin π
(2) π2 = π₯2 + π¦2 , tan π =π¦
π₯ ππ π₯ β 0
Changing polar coordinates into rectangular coordinates
#4 (a) (5,5π
6) (b) (β6,
7π
3)
#8 (10, arccos (β1
3))
Change the rectangular coordinates to polar
coordinates with π > 0 πππ 0 β€ π β€ 2π
#10 (a) (3β3, 3)
#12 (b) (β4, 4β3)
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Solving polar equations
We use the relationship between rectangular and polar coordinates (π₯ = π πππ π, π¦ = π sin π ) to
transform an equation in x and y into a polar equation.
Find a polar equation that has the same graph as the equation in π₯ and π¦.
#14 π¦ = 2
#18 π₯2 + π¦2 = 2
#20 π₯2 = 8π¦
#24 2π¦ = βπ₯ + 4
#30 π₯2 β π¦2 = 9
#34 π₯2 + (π¦ β 1)2 = 1
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