16. Conic Sections Concept Overview
CONIC SECTIONS | CONCEPT OVERVIEW The topic of CONIC SECTIONS can be referenced on pages 26-27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: An equation with second-order terms such at those in a quadratic function will produce a CONIC SECTION when graphed on a Cartesian coordinate system. The FORMULA FOR THE GENERAL FORM OF THE CONIC SECTION EQUATION can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The general quadratic form of a conic section is given by the standard equation: π΄π₯ # + π΅π₯π¦ + πΆπ¦ # + π·π₯ + πΈπ¦ + πΉ = 0 Where not both π΄ πππ πΆ are zero The FORMULAS TO CLASSIFY THE TYPE OF CONIC SECTION can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We can define the type of conic section by using the discriminant to classify the properties of the function:
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β’ If π΅ # β 4π΄πΆ < 0, the conic section is an ELLIPSE β’ If π΅ # β 4π΄πΆ > 0, the conic section is a HYPERBOLA. β’ If π΅ # β 4π΄πΆ = 0, the conic section is a PARABOLA β’ If π΄ = πΆ and π΅ = 0, the conic section is a CIRCLE. β’ If π΄ = π΅ = πΆ = 0, the function defines a STRAIGHT LINE. The NORMAL FORM OF THE CONIC SECTION EQUATION is: π₯ # + π¦ # + 2ππ₯ + 2ππ¦ + π = 0 Where: β’ The conic section has a principal axis parallel to a coordinate system β’ β = βπ β’ π = βπ β’ π = π# + π# β π o If π# + π # β π > 0, a CIRCLE with center βπ, βπ is defined o If π# + π # β π = 0, a POINT at coordinates (βπ, βπ) is defined o If π# + π # β π < 0, the LOCUS is IMAGINARY. The FORMULA FOR THE ECCENTRICITY OF A CONIC SECTION can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The ECCENTRICITY of a conic section is calculated as:
π = πΈπΆπΆπΈπππ πΌπΆπΌππ =
DEF G DEF H
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CONCEPT EXAMPLE: Class the graph of the general quadratic equation given by the function: 4π₯ # β 4π₯π¦ + π¦ # β 5 5π₯ + 5 = 0 A. πΈπππππ π B. π»π¦πππππππ C. πΆπππππ D. ππππππππa
SOLUTION: The FORMULA FOR THE GENERAL FORM OF THE CONIC SECTION EQUATION can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The general quadratic form of a conic section is given by the standard equation: π΄π₯ # + π΅π₯π¦ + πΆπ¦ # + π·π₯ + πΈπ¦ + πΉ = 0 In the general quadratic equation 4π₯ # β 4π₯π¦ + π¦ # β 5 5π₯ + 5 = 0, we find that: π΄ = 4, π΅ = β4, πππ πΆ = 1 The FORMULAS TO CLASSIFY THE TYPE OF CONIC SECTION can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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We then calculate the discriminant to classify the conic section: π΅ # β 4π΄πΆ = β4
#
β4 4 1 =0
We know that when the πππ πππππππππ‘ = 0, the conic section is a ππππππππ.
Therefore, the correct answer choice is D. π·πππππππ
CASE π: PARABOLAS π = π The TOPIC OF PARABOLAS can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A PARABOLA is the set of all points in a plane that are equidistant from a fixed line called the DIRECTRIX and a fixed point not on the line called the FOCUS. The most general form of a quadratic function is: π π₯ = ππ₯ # + ππ₯ + π The graph of a quadratic equation is a parabola and generally takes the shape of a βUβ. Every parabola has an imaginary line that runs down the center of it called the axis of symmetry; where each side is a mirror image of the other. If one point of the parabola is known, then the point directly on the other side is known based on the axis of symmetry.
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The highest or lowest point of a parabola is called the VERTEX.
The FORMULA FOR THE STANDARD FORM OF A PARABOLA can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The standard form of the equation for a parabola is represented by the equation: π¦βπ
#
= 2π(π₯ β β)
Where: β’ The center of the parabola has coordinates β, π β’ The focus of the parabola has coordinates: (π/2,0) β’ The directrix of the parabola has coordinate: π₯ = β
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The FORMULAS FOR THE VERTICAL AND HORIZONTAL AXES OF SYMMETRY are not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. A parabola that is centered at the origin β, π = (0,0) and has a VERTICAL AXIS OF SYMMETRY is represented by the equation: π₯ # = 4ππ¦ Where: β’ If π > 0, the parabola opens UPWARD β’ If π < 0, the parabola opens DOWNWARD.
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A parabola that is centered at the origin β, π = (0,0) has a HORIZONTAL AXIS OF SYMMETRY is represented by the equation: π¦ # = 4ππ₯ Where: β’ If π > 0, the parabola opens RIGHT β’ If π < 0, the parabola opens LEFT
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CONCEPT EXAMPLE: Determine the general quadratic equation of the parabola with a vertex at the origin and focus at (β2, 0). 4 π₯+1
#
+ π¦+3
#
=1
A. π¦ # + 3π₯ = 0 B. π¦ # + 5π₯ = 0 C. π¦ # + 8π₯ = 0 D. π¦ # + 13π₯ = 0
SOLUTION: The TOPIC OF PARABOLAS can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A parabola with vertex at the origin and focus on the x-axis has form π¦ # = 4ππ₯, with (π, 0) as the coordinates of the focus. Since the x-coordinate of the focus given is β2, the general quadratic equation of this parabola is found by replacing βπβ with β2 in the equation π¦ # = 4ππ₯. In the standard quadratic form, we find: π¦ # = β8π₯ or π¦ # + 8π₯ = 0
Therefore, the correct answer choice is C. ππ + ππ = π
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CASE π: ELLIPSES π < π The TOPIC OF ELLIPSES can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. An ELLIPSE is a set of all points in a plane, the sum of whose distances from two fixed points, called βFOCI,β is a constant that is equal to the length of the major axis of the ellipse.
The FORMULA FOR THE STANDARD FORM OF AN ELLIPSE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing The standard form of the equation for an ellipse is represented by the equation: π₯ββ π#
#
π¦βπ + π#
#
=1
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Where: β’ The center of the ellipse has coordinates β, π β’ The variable βπβ is the square root of the number under the βπ₯β term and is the amount that we move to right or left from the center. β’ The variable βπβ is the square root of the number under the βπ¦β term and is the amount we move up or down from the center. An ellipse can be quickly graphed by defining the four extreme points: β’ Right Most Point: β + π, π β’ Left Most Point: (β β π, π) β’ Top Most Point: (β, π + π) β’ Bottom Most Point: (β, π β π) The FORMULA FOR THE ECCENTRICITY OF AN ELLIPSE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing The eccentricity of an ellipse is represented by the equation:
π=
π# π 1β # = π π
π = π 1 β π#
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The FORMULAS FOR FOCUS AND DIRECTRIX OF AN ELLIPSE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing The focus of an ellipse is represented by the coordinates: (Β±ππ, 0) The directrix of an ellipse is represented by the equation: π₯ = Β±
j k
An ellipse that is centered at the origin β, π = (0,0) and has vertices on the x-axis is represented by the equation: π₯# π¦# + =1 π# π# An ellipse that is centered at the origin β, π = (0,0) and has vertices on the y-axis is represented by the equation: π¦# π₯# + =1 π# π#
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Ellipse Type 1: Horizontal Orientation The FORMULA FOR THE STANDARD FORM OF AN ELLIPSE WITH A HORIZONTAL ORIENTATION IS not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. An ellipse with a horizontal orientation that has the standard form such that the variable βπβ is under the π₯ β π‘πππ of the equation, is shown in the expression: π₯ββ π#
#
π¦βπ + π#
#
=1
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Ellipse Type 2: Vertical Orientation Orientation The FORMULA FOR THE STANDARD FORM OF AN ELLIPSE WITH A VERTICAL ORIENTATION IS not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. An ellipse with a vertical orientation that has the standard form such that the variable βπβ is under the π¦ β π‘πππ of the equation, is shown in the expression: π¦ββ π#
#
π₯βπ + π#
#
=1
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CONCEPT EXAMPLE: What is the center and right most point of the ellipse given the function: 4 π₯+1
#
#
+ π¦+3
=1 p
A. πΆπππ‘ππ πΆππππππππ‘ππ : β1, 3 ; π ππβπ‘ πππ π‘ πΆππππππππ‘ππ : (β , β3) #
p
B. πΆπππ‘ππ πΆππππππππ‘ππ : 1, 3 ; π ππβπ‘ πππ π‘ πΆππππππππ‘ππ : ( , β3) #
p
C. πΆπππ‘ππ πΆππππππππ‘ππ : 1, β 3 ; π ππβπ‘ πππ π‘ πΆππππππππ‘ππ : (β , β3) #
D. πΆπππ‘ππ πΆππππππππ‘ππ : 1, β3 ; π ππβπ‘ πππ π‘ πΆππππππππ‘ππ :
p #
, β3
SOLUTION: The FORMULA FOR THE STANDARD FORM OF AN ELLIPSE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing Recall that the standard form of an ellipse is given by the equation: π₯ββ π#
#
π¦βπ + π#
#
=1
The first step is to rearrange the equation that is given so that it is in a more familiar form that we are able to work with. Doing so we get: π₯+1 1 4
#
+ π¦+3
#
=1
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Note that there isnβt a variable below the y term, which indicates that π # = 1. This value can be placed in to the equation so that it replicates the standard form: π₯+1 1 4
#
+
π¦+3 1
#
=1
In the standard form, the point (β, π) is the center of the ellipse. Therefore, the center of the ellipse is given by the coordinates (β, π) = (β1,3) We can now calculate the values of π πππ π that will let us determine the extrema of the ellipse:
π=
1 πππ π = 1 2
The right most point is given by the coordinates 1 1 β + π, π = β1 + , 3 = β , β3 2 2
Therefore, the correct answer choice is π
A. πͺπππππ πͺππππ ππππππ: βπ, π ; πΉππππ π΄πππ πͺππππ ππππππ: (β , βπ) π
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CASE π HYPERBOLA: π > π The TOPIC OF HYPERBOLAS can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A HYPERBOLA is the set of all points in a plane, the difference of whose distance from two fixed point (foci) is a constant that is equal to the length of the transverse axis of the hyperbola.
The FORMULA FOR THE STANDARD FORM OF A HYPERBOLA can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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The standard form of the equation for a HYPERBOLA is represented by the equation: π₯ββ π#
#
π¦βπ β π#
#
=1
Where: β’ The center of the ellipse has coordinates β, π The FORMULA FOR THE ECCENTRICITY OF A HYPERBOLA can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The eccentricity of a hyperbola is represented by the equation:
π=
π# π 1+ # = π π
π = π π# β 1 The FORMULAS FOR THE FOCUS AND DIRECTRIX OF A HYPERBOLA can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The focus of an ellipse is represented by the coordinates: (Β±ππ, 0)
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The directrix of an ellipse is represented by the equation:
π₯=Β±
π π
An hyperbola that is centered at the origin β, π = (0,0) and has vertices on the x-axis is represented by the equation: π₯# π¦# β =1 π# π# A hyperbola that is centered at the origin β, π = (0,0), and has vertices on the y-axis, is represented by the equation: π¦# π₯# β =1 π# π# The graph of the equation π₯π¦ = π (π ππ π ππππ π‘πππ‘) is also a hyperbola. Its center is at the origin, and its asymptopes are the π₯ β and π¦ β axes.
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There are two standard forms for the hyperbola defined by the following characteristics: Standard Form
(π₯ β β)# (π¦ β π)# β β1 π# π#
(π¦ β π)# (π₯ β β)# β β1 π# π#
Center
(β, π)
(β, π)
Orientation
πππππ ππππ‘ πππ πππβπ‘
πππππ π’π πππ πππ€π
Vertices
(β + π, π)πππ (β β π, π)
Slope of Asymptotes Equations of Asymptotes
Β±
(β, π + π)πππ (β, π β π)
π π
π π¦ = π Β± (π₯ β β) π
Β±
π π
π π¦ = π Β± (π₯ β β) π
Hyperbolas are two parabola-like shaped pieces that open either up and down or left and right. Just like parabolas, each of the pieces has a vertex. There are also two lines that may be illustrated along with the Hyperbola, these are called asymptotes. The asymptotes are not officially part of the the hyperbola. The point where the two asymptotes cross is called the center of the hyperbola
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CONCEPT EXAMPLE: Determine the general quadratic equation of a hyperbola with asymptotes along the π₯ β πππ π¦ β ππ₯ππ and passes through point (3,2). A. 2π₯π¦ β 2 = 0 B. π₯π¦ β 5 = 0 C. π₯π¦ β 6 = 0 D. π₯π¦ β 12 = 0
SOLUTION: The TOPIC OF HYPERBOLAS can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A hyperbola with asymptotes along the x- and y- axes has the form π₯π¦ = π. Since the hyperbola passes through point (3,2), π = π₯π¦ = 3 2 = 6. The general quadratic equation of this hyperbola is found by replacing βπβ in the equation π₯π¦ = π: π₯π¦ = 6 or π₯π¦ β 6 = 0
Therefore, the correct answer choice is C. ππ β π = π
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CASE π CIRCLES: π = π The TOPIC OF HYPERBOLAS can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A CIRCLE is a set of all points in a plane that lie a fixed distance (radius) from a defined central point (β, π)
The FORMULA FOR THE STANDARD FORM A CIRCLE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The STANDARD FORM OF A CIRCLE centered at coordinates (β, π) with radius βπβ is represented by the equation:
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π=
π₯ββ
#
+ π¦βπ
#
π₯ββ
or
#
+ π¦βπ
#
= π#
A TANGENT to a circle is a line with an endpoint not within the circle that intersects the circle in one and only one point. Every tangent to a circle is perpendicular to a radius drawn to the point of tangency. The FORMULA FOR THE LENGTH OF A TANGENT LINE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The length of the tangent line from a point on a circle to a point (π₯ β , π¦β²) is given by the equation: π‘# = π₯β β β
#
+ π¦β β π
#
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CONCEPT EXAMPLE: Determine the equation of the circle with a radius of 6 and centered at (β3,4). A. π₯ + 3
#
+ π¦β4
#
= 36
B. π₯ + 1
#
+ π¦β6
#
= 49
C. π₯ β 1
#
+ π¦+4
#
= 74
D. π₯ β 5
#
+ π¦+7
#
= 81
SOLUTION: The FORMULA FOR THE STANDARD FORM OF A CIRCLE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The STANDARD FORM OF A CIRCLE centered at coordinates (β, π) with radius βπβ is represented by the equation:
π=
π₯ββ
#
+ π¦βπ
#
or
π₯ββ
#
+ π¦βπ
#
= π#
The center is given as (β, π) = (β3, 4) with a radius of π = 6: Plugging these values into the standard equation provides the result: π₯+3
#
+ π¦β4
#
= 36
Therefore, the correct answer choice is A. π + π
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+ πβπ
π
= ππ
CONCEPT INTRO: An equation with second-order terms such at those in a quadratic function will produce a CONIC SECTION when graphed on a Cartesian coordinate system. The FORMULA FOR THE GENERAL FORM OF THE CONIC SECTION EQUATION can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The general quadratic form of a conic section is given by the standard equation: π΄π₯ # + π΅π₯π¦ + πΆπ¦ # + π·π₯ + πΈπ¦ + πΉ = 0 Where not both π΄ πππ πΆ are zero The FORMULAS TO CLASSIFY THE TYPE OF CONIC SECTION can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We can define the type of conic section by using the discriminant to classify the properties of the function:
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β’ If π΅ # β 4π΄πΆ < 0, the conic section is an ELLIPSE β’ If π΅ # β 4π΄πΆ > 0, the conic section is a HYPERBOLA. β’ If π΅ # β 4π΄πΆ = 0, the conic section is a PARABOLA β’ If π΄ = πΆ and π΅ = 0, the conic section is a CIRCLE. β’ If π΄ = π΅ = πΆ = 0, the function defines a STRAIGHT LINE. The NORMAL FORM OF THE CONIC SECTION EQUATION is: π₯ # + π¦ # + 2ππ₯ + 2ππ¦ + π = 0 Where: β’ The conic section has a principal axis parallel to a coordinate system β’ β = βπ β’ π = βπ β’ π = π# + π# β π o If π# + π # β π > 0, a CIRCLE with center βπ, βπ is defined o If π# + π # β π = 0, a POINT at coordinates (βπ, βπ) is defined o If π# + π # β π < 0, the LOCUS is IMAGINARY. The FORMULA FOR THE ECCENTRICITY OF A CONIC SECTION can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The ECCENTRICITY of a conic section is calculated as:
π = πΈπΆπΆπΈπππ πΌπΆπΌππ =
DEF G DEF H
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CONCEPT EXAMPLE: Class the graph of the general quadratic equation given by the function: 4π₯ # β 4π₯π¦ + π¦ # β 5 5π₯ + 5 = 0 A. πΈπππππ π B. π»π¦πππππππ C. πΆπππππ D. ππππππππa
SOLUTION: The FORMULA FOR THE GENERAL FORM OF THE CONIC SECTION EQUATION can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The general quadratic form of a conic section is given by the standard equation: π΄π₯ # + π΅π₯π¦ + πΆπ¦ # + π·π₯ + πΈπ¦ + πΉ = 0 In the general quadratic equation 4π₯ # β 4π₯π¦ + π¦ # β 5 5π₯ + 5 = 0, we find that: π΄ = 4, π΅ = β4, πππ πΆ = 1 The FORMULAS TO CLASSIFY THE TYPE OF CONIC SECTION can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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We then calculate the discriminant to classify the conic section: π΅ # β 4π΄πΆ = β4
#
β4 4 1 =0
We know that when the πππ πππππππππ‘ = 0, the conic section is a ππππππππ.
Therefore, the correct answer choice is D. π·πππππππ
CASE π: PARABOLAS π = π The TOPIC OF PARABOLAS can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A PARABOLA is the set of all points in a plane that are equidistant from a fixed line called the DIRECTRIX and a fixed point not on the line called the FOCUS. The most general form of a quadratic function is: π π₯ = ππ₯ # + ππ₯ + π The graph of a quadratic equation is a parabola and generally takes the shape of a βUβ. Every parabola has an imaginary line that runs down the center of it called the axis of symmetry; where each side is a mirror image of the other. If one point of the parabola is known, then the point directly on the other side is known based on the axis of symmetry.
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The highest or lowest point of a parabola is called the VERTEX.
The FORMULA FOR THE STANDARD FORM OF A PARABOLA can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The standard form of the equation for a parabola is represented by the equation: π¦βπ
#
= 2π(π₯ β β)
Where: β’ The center of the parabola has coordinates β, π β’ The focus of the parabola has coordinates: (π/2,0) β’ The directrix of the parabola has coordinate: π₯ = β
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The FORMULAS FOR THE VERTICAL AND HORIZONTAL AXES OF SYMMETRY are not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. A parabola that is centered at the origin β, π = (0,0) and has a VERTICAL AXIS OF SYMMETRY is represented by the equation: π₯ # = 4ππ¦ Where: β’ If π > 0, the parabola opens UPWARD β’ If π < 0, the parabola opens DOWNWARD.
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A parabola that is centered at the origin β, π = (0,0) has a HORIZONTAL AXIS OF SYMMETRY is represented by the equation: π¦ # = 4ππ₯ Where: β’ If π > 0, the parabola opens RIGHT β’ If π < 0, the parabola opens LEFT
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CONCEPT EXAMPLE: Determine the general quadratic equation of the parabola with a vertex at the origin and focus at (β2, 0). 4 π₯+1
#
+ π¦+3
#
=1
A. π¦ # + 3π₯ = 0 B. π¦ # + 5π₯ = 0 C. π¦ # + 8π₯ = 0 D. π¦ # + 13π₯ = 0
SOLUTION: The TOPIC OF PARABOLAS can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A parabola with vertex at the origin and focus on the x-axis has form π¦ # = 4ππ₯, with (π, 0) as the coordinates of the focus. Since the x-coordinate of the focus given is β2, the general quadratic equation of this parabola is found by replacing βπβ with β2 in the equation π¦ # = 4ππ₯. In the standard quadratic form, we find: π¦ # = β8π₯ or π¦ # + 8π₯ = 0
Therefore, the correct answer choice is C. ππ + ππ = π
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CASE π: ELLIPSES π < π The TOPIC OF ELLIPSES can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. An ELLIPSE is a set of all points in a plane, the sum of whose distances from two fixed points, called βFOCI,β is a constant that is equal to the length of the major axis of the ellipse.
The FORMULA FOR THE STANDARD FORM OF AN ELLIPSE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing The standard form of the equation for an ellipse is represented by the equation: π₯ββ π#
#
π¦βπ + π#
#
=1
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Where: β’ The center of the ellipse has coordinates β, π β’ The variable βπβ is the square root of the number under the βπ₯β term and is the amount that we move to right or left from the center. β’ The variable βπβ is the square root of the number under the βπ¦β term and is the amount we move up or down from the center. An ellipse can be quickly graphed by defining the four extreme points: β’ Right Most Point: β + π, π β’ Left Most Point: (β β π, π) β’ Top Most Point: (β, π + π) β’ Bottom Most Point: (β, π β π) The FORMULA FOR THE ECCENTRICITY OF AN ELLIPSE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing The eccentricity of an ellipse is represented by the equation:
π=
π# π 1β # = π π
π = π 1 β π#
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The FORMULAS FOR FOCUS AND DIRECTRIX OF AN ELLIPSE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing The focus of an ellipse is represented by the coordinates: (Β±ππ, 0) The directrix of an ellipse is represented by the equation: π₯ = Β±
j k
An ellipse that is centered at the origin β, π = (0,0) and has vertices on the x-axis is represented by the equation: π₯# π¦# + =1 π# π# An ellipse that is centered at the origin β, π = (0,0) and has vertices on the y-axis is represented by the equation: π¦# π₯# + =1 π# π#
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Ellipse Type 1: Horizontal Orientation The FORMULA FOR THE STANDARD FORM OF AN ELLIPSE WITH A HORIZONTAL ORIENTATION IS not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. An ellipse with a horizontal orientation that has the standard form such that the variable βπβ is under the π₯ β π‘πππ of the equation, is shown in the expression: π₯ββ π#
#
π¦βπ + π#
#
=1
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Ellipse Type 2: Vertical Orientation Orientation The FORMULA FOR THE STANDARD FORM OF AN ELLIPSE WITH A VERTICAL ORIENTATION IS not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. An ellipse with a vertical orientation that has the standard form such that the variable βπβ is under the π¦ β π‘πππ of the equation, is shown in the expression: π¦ββ π#
#
π₯βπ + π#
#
=1
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CONCEPT EXAMPLE: What is the center and right most point of the ellipse given the function: 4 π₯+1
#
#
+ π¦+3
=1 p
A. πΆπππ‘ππ πΆππππππππ‘ππ : β1, 3 ; π ππβπ‘ πππ π‘ πΆππππππππ‘ππ : (β , β3) #
p
B. πΆπππ‘ππ πΆππππππππ‘ππ : 1, 3 ; π ππβπ‘ πππ π‘ πΆππππππππ‘ππ : ( , β3) #
p
C. πΆπππ‘ππ πΆππππππππ‘ππ : 1, β 3 ; π ππβπ‘ πππ π‘ πΆππππππππ‘ππ : (β , β3) #
D. πΆπππ‘ππ πΆππππππππ‘ππ : 1, β3 ; π ππβπ‘ πππ π‘ πΆππππππππ‘ππ :
p #
, β3
SOLUTION: The FORMULA FOR THE STANDARD FORM OF AN ELLIPSE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing Recall that the standard form of an ellipse is given by the equation: π₯ββ π#
#
π¦βπ + π#
#
=1
The first step is to rearrange the equation that is given so that it is in a more familiar form that we are able to work with. Doing so we get: π₯+1 1 4
#
+ π¦+3
#
=1
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Note that there isnβt a variable below the y term, which indicates that π # = 1. This value can be placed in to the equation so that it replicates the standard form: π₯+1 1 4
#
+
π¦+3 1
#
=1
In the standard form, the point (β, π) is the center of the ellipse. Therefore, the center of the ellipse is given by the coordinates (β, π) = (β1,3) We can now calculate the values of π πππ π that will let us determine the extrema of the ellipse:
π=
1 πππ π = 1 2
The right most point is given by the coordinates 1 1 β + π, π = β1 + , 3 = β , β3 2 2
Therefore, the correct answer choice is π
A. πͺπππππ πͺππππ ππππππ: βπ, π ; πΉππππ π΄πππ πͺππππ ππππππ: (β , βπ) π
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CASE π HYPERBOLA: π > π The TOPIC OF HYPERBOLAS can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A HYPERBOLA is the set of all points in a plane, the difference of whose distance from two fixed point (foci) is a constant that is equal to the length of the transverse axis of the hyperbola.
The FORMULA FOR THE STANDARD FORM OF A HYPERBOLA can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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The standard form of the equation for a HYPERBOLA is represented by the equation: π₯ββ π#
#
π¦βπ β π#
#
=1
Where: β’ The center of the ellipse has coordinates β, π The FORMULA FOR THE ECCENTRICITY OF A HYPERBOLA can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The eccentricity of a hyperbola is represented by the equation:
π=
π# π 1+ # = π π
π = π π# β 1 The FORMULAS FOR THE FOCUS AND DIRECTRIX OF A HYPERBOLA can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The focus of an ellipse is represented by the coordinates: (Β±ππ, 0)
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The directrix of an ellipse is represented by the equation:
π₯=Β±
π π
An hyperbola that is centered at the origin β, π = (0,0) and has vertices on the x-axis is represented by the equation: π₯# π¦# β =1 π# π# A hyperbola that is centered at the origin β, π = (0,0), and has vertices on the y-axis, is represented by the equation: π¦# π₯# β =1 π# π# The graph of the equation π₯π¦ = π (π ππ π ππππ π‘πππ‘) is also a hyperbola. Its center is at the origin, and its asymptopes are the π₯ β and π¦ β axes.
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There are two standard forms for the hyperbola defined by the following characteristics: Standard Form
(π₯ β β)# (π¦ β π)# β β1 π# π#
(π¦ β π)# (π₯ β β)# β β1 π# π#
Center
(β, π)
(β, π)
Orientation
πππππ ππππ‘ πππ πππβπ‘
πππππ π’π πππ πππ€π
Vertices
(β + π, π)πππ (β β π, π)
Slope of Asymptotes Equations of Asymptotes
Β±
(β, π + π)πππ (β, π β π)
π π
π π¦ = π Β± (π₯ β β) π
Β±
π π
π π¦ = π Β± (π₯ β β) π
Hyperbolas are two parabola-like shaped pieces that open either up and down or left and right. Just like parabolas, each of the pieces has a vertex. There are also two lines that may be illustrated along with the Hyperbola, these are called asymptotes. The asymptotes are not officially part of the the hyperbola. The point where the two asymptotes cross is called the center of the hyperbola
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CONCEPT EXAMPLE: Determine the general quadratic equation of a hyperbola with asymptotes along the π₯ β πππ π¦ β ππ₯ππ and passes through point (3,2). A. 2π₯π¦ β 2 = 0 B. π₯π¦ β 5 = 0 C. π₯π¦ β 6 = 0 D. π₯π¦ β 12 = 0
SOLUTION: The TOPIC OF HYPERBOLAS can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A hyperbola with asymptotes along the x- and y- axes has the form π₯π¦ = π. Since the hyperbola passes through point (3,2), π = π₯π¦ = 3 2 = 6. The general quadratic equation of this hyperbola is found by replacing βπβ in the equation π₯π¦ = π: π₯π¦ = 6 or π₯π¦ β 6 = 0
Therefore, the correct answer choice is C. ππ β π = π
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CASE π CIRCLES: π = π The TOPIC OF HYPERBOLAS can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A CIRCLE is a set of all points in a plane that lie a fixed distance (radius) from a defined central point (β, π)
The FORMULA FOR THE STANDARD FORM A CIRCLE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The STANDARD FORM OF A CIRCLE centered at coordinates (β, π) with radius βπβ is represented by the equation:
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π=
π₯ββ
#
+ π¦βπ
#
π₯ββ
or
#
+ π¦βπ
#
= π#
A TANGENT to a circle is a line with an endpoint not within the circle that intersects the circle in one and only one point. Every tangent to a circle is perpendicular to a radius drawn to the point of tangency. The FORMULA FOR THE LENGTH OF A TANGENT LINE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The length of the tangent line from a point on a circle to a point (π₯ β , π¦β²) is given by the equation: π‘# = π₯β β β
#
+ π¦β β π
#
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CONCEPT EXAMPLE: Determine the equation of the circle with a radius of 6 and centered at (β3,4). A. π₯ + 3
#
+ π¦β4
#
= 36
B. π₯ + 1
#
+ π¦β6
#
= 49
C. π₯ β 1
#
+ π¦+4
#
= 74
D. π₯ β 5
#
+ π¦+7
#
= 81
SOLUTION: The FORMULA FOR THE STANDARD FORM OF A CIRCLE can be referenced under the topic of CONIC SECTIONS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The STANDARD FORM OF A CIRCLE centered at coordinates (β, π) with radius βπβ is represented by the equation:
π=
π₯ββ
#
+ π¦βπ
#
or
π₯ββ
#
+ π¦βπ
#
= π#
The center is given as (β, π) = (β3, 4) with a radius of π = 6: Plugging these values into the standard equation provides the result: π₯+3
#
+ π¦β4
#
= 36
Therefore, the correct answer choice is A. π + π
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+ πβπ
π
= ππ
