32.1 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 shown in quadratic form will produce a CONIC SECTION when graphed in a Cartesian coordinate system. The GENERAL FORM of a 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 written as: Ax # + Bxy + Cy # + Dx + Ey + F = 0 Where both π΄ πππ πΆ are real, non-zero numbers. The RULES to help us 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 can quickly classify the type of conic section using the discriminant, such that: 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 written as: 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.
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The ECCENTRICITY of a conic section is defined as:
π = πΈπΆπΆπΈπππ πΌπΆπΌππ =
cos π cos π
CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Classify the shape of the graph given the general quadratic equation: 4π₯ # β 4π₯π¦ + π¦ # β 5 5π₯ + 5 = 0 A. Ellipse B. Hyperbola C. Circle D. Parabola
SOLUTION: The GENERAL FORM of a 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.
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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 Letβs work the formula we are given so that it is presented to us in this general form. In this problem, we are given: 4π₯ # β 4π₯π¦ + π¦ # β 5 5π₯ + 5 = 0 Referring to the general quadratic form of a conic section, it appears everything is already set up as we would want it. With that being established, we find that: A=4 B = -4 C=1 The RULES to help us 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. To classify the type of conic section, we must first calculate the DISCRIMINANT given the data from the formula stated.
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The DISCRIMINANT can be defined using the formula: π΅ # β 4π΄πΆ With all the proper variables define, we simply just need to plug and play, giving us: π΅ # β 4π΄πΆ = β4
#
β4 4 1 =0
With the DISCRIMINANT defined, we refer to the set of RULES stated for us in the NCEES REFERENCE HANDBOOK, which state that: 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. With a DISCRIMANT of zero, we can conclude that the graph of the CONIC SECTION will be the shape of a PARABOLA. The correct answer choice is D. π·πππππππ CASE π: PARABOLAS π = π Information revolving around 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. Made with
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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 can be written as: π π₯ = ππ₯ # + ππ₯ + π When plotted out, the graph of a quadratic function 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. On either side of this line, the parabola is a one to one mirror image of itself. Therefore, if one point on a parabola is known, then the point directly on the other side is known as well, based on the AXIS OF SYMMETRY. The highest or lowest point of a parabola is called the VERTEX. The GENBERAL FORMULA in 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)
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Where: β’ The center of the parabola is located at the coordinates β, π β’ The focus of the parabola is located at runs along the coordinate: π₯ = β
f #
Given this general formula, a parabola can be plotted as such:
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. Therefore, we must take note to retain 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:
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π₯ # = 4ππ¦ Where: β’ If π > 0, the parabola opens UPWARD β’ If π < 0, the parabola opens DOWNWARD. Illustrating this, we have:
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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Generally illustrating this, we have:
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CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
The general quadratic equation of a parabola with a vertex at the origin and focus located at (β2, 0) is best represented as: A. π¦ # + 3π₯ = 0 B. π¦ # + 5π₯ = 0 C. π¦ # + 8π₯ = 0 D. π¦ # + 13π₯ = 0
SOLUTION: Information revolving around 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 a vertex at the origin and a focus on the x-axis takes the general form: π¦ # = 4ππ₯ Where (π, 0) is the coordinates of the focus.
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Since the x-coordinate of the focus is defined as β2, which is given in our problem statement, the general quadratic equation of this parabola can be defined by replacing βπβ with β2 such that: π¦ # = 4(β2)π₯ Which gives us: π¦ # = β8π₯ Rearranging this formula so that it is in standard quadratic form, we find: π¦ # + 8π₯ = 0 The correct answer choice is C. ππ + ππ = π
CASE π: ELLIPSES π < π Information revolving around 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 lays out geometrically as an βOvalβ when presented in your typical x-y coordinate system.
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Every ELLIPSE has two fixed points called the FOCI, wherein which, taking any single point on this ellipse, the sum of the lengths from that point to each of the FOCI remains CONSTANT. This characteristic is what drives the shape of the ELLIPSE. This CONSTANT summation of lengths is also the length of the MAJOR AXIS for an ELLIPSE. An ELLIPSE can be generally illustrated as:
The FORMULA for an ELLIPSE in STANDARD FORM 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 for an ellipse is represented by the equation: π₯ββ π#
#
π¦βπ + π#
#
=1
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Where: β’ The the coordinates β, π represents that center of the ellipse β’ The variable βπβ gives us the distance from the center x-coordinate to the right and left most point of the ELLIPSE β’ The variable βπβ gives us the distance from the center y-coordinate to the upper and lower most point of the ELLIPSE An ELLIPSE can quickly be plotted after defining the four extreme points, such that: 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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Where: β’ βaβ is the distance from the center of the ELLIPSE to the LEFT or RIGHT most point β’ βbβ is the distance from the center of the ELLIPSE to the UPPER or LOWER most point β’ βcβ is the FIXED DISTANCE from the center of the ELLIPSE to either of the FOCI The FORMULAS for locating the 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±
π π
An ELLIPSE that is centered at the origin β, π = (0,0) and has itβs FOCI on the x-axis is represented by the equation: π₯# π¦# + =1 π# π#
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An ELLIPSE that is centered at the origin β, π = (0,0) and has itβs FOCI on the y-axis is represented by the equation: π¦# π₯# + =1 π# π# ELLIPSE TYPE 1: HORIZONTAL ORIENTATION The FORMULA in 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 is represented in standard form as: π₯ββ π#
#
π¦βπ + π#
#
=1
An ELLIPSE with a HORIZONTAL ORIENTATION can generally be drawn as:
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ELLIPSE TYPE 2: VERTICAL ORIENTATION The FORMULA in 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. Typically, we will see that variable βaβ tied directly to the x-term in the general form of an ELLIPSEβ¦resulting in an ellipse with a HORIZONTAL ORIENTATION as described above. However, when βaβ is tied directly to the y-term, then the ORIENTATION of the ELLIPSE flips, presenting itself VERTICALLY. The general form of an ELLIPSE with a VERTICAL ORIENTATION can be written as: π¦ββ π#
#
π₯βπ + π#
#
=1
An ELLIPSE with a VERTICAL ORIENTATION can generally be drawn as:
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CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
The center of the ellipse driven by the defined function is most close to: 4 π₯+1
#
+ π¦β3
#
=1
A. β1, 3 B. 1, 3 C. 1, β 3 D. β1, β3
SOLUTION: The FORMULA of an ELLIPSE in STANDARD FORM 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
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The first step is to rearrange the formula that we are given in to a form that best represents this standard form. Doing so we get: π₯+1 1 4
#
+ π¦β3
#
=1
Note that there isnβt a variable below the y term, which indicates that π # = 1. We can place this value in to the equation so that it replicates the standard form: π₯+1 1 4
#
π¦β3 + 1
#
=1
In the standard form, the point (β, π) is the center of the ellipse. Revisiting where these variables fall within the formula, the standard formula reads: π₯ββ π#
#
π¦βπ + π#
#
=1
Therefore, the center of the ellipse is given by the coordinates (β, π) which are: (β1,3) The correct answer choice is A. βπ, π
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CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
The right most point of the ellipse driven by the defined function is most close to: 4 π₯+1
#
+ π¦β3
#
=1
s
A. (β , 3) #
s
B. ( , β3) #
s
C. (β , β3) #
D.
s #
,3
SOLUTION: The FORMULA of an ELLIPSE in STANDARD FORM 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 can quickly be plotted after defining the four extreme points, such that: β’ Right Most Point: β + π, π β’ Left Most Point: (β β π, π)
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β’ Top Most Point: (β, π + π) β’ Bottom Most Point: (β, π β π) In this problem, we are only concerned with the RIGHT MOST POINT: β’ Right Most Point: β + π, π We wonβt be able to determine this point without definition on where the center of this ellipse falls, lets get this point. 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 formula that we are given in to a form that best represents this standard form. Doing so we get: π₯+1 1 4
#
+ π¦β3
#
=1
Note that there isnβt a variable below the y term, which indicates that π # = 1.
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We can place this value in to the equation so that it replicates the standard form: π₯+1 1 4
#
π¦β3 + 1
#
=1
In the standard form, the point (β, π) is the center of the ellipse. Revisiting where these variables fall within the formula, the standard formula reads: π₯ββ π#
#
π¦βπ + π#
#
=1
Therefore, the center of the ellipse is given by the coordinates β, π which are: (β1,3) We can now define the values of the variable π as it is represented in this formula, we are told that: 1 π# = 4 Which makes: 1 π= 2
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The right most point is given by the coordinates: 1 β + π, π = β1 + , 3 2 Or: 1 β ,3 2 π
The correct answer choice is A. (β , π) π
CASE π HYPERBOLA: π > π Information revolving around 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 lays out geometrically as two reverse facing parabolas in typical x-y coordinate system. Every HYPERBOLA has two fixed points called the FOCI, wherein which, taking any single point on this hyperbola, the difference of the lengths from that point to each of the FOCI remains CONSTANT. This characteristic is what drives the shape of the HYPERBOLA.
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This CONSTANT difference of lengths is also the length of the TRANSVERSE AXIS for the HYPERBOLA. A HYPERBOLA can be generally illustrated as:
The FORMULA for a HYPERBOLA in the STANDARD FORM 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 HYPERBOLA is written as: π₯ββ π#
#
π¦βπ β π#
#
=1
Where: β’ The center of the HYPERBOLA is located at the coordinates β, π
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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 to define 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 a HYPERBOLA is represented by the coordinates: (Β±ππ, 0) The DIRECTRIX of a HYPERBOLA is represented by the equation:
π₯=Β±
π π
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A HYPERBOLA that is centered at the origin β, π = (0,0) and has a HYPERBOLA 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 FOCI on the y-axis, is represented by the equation: π¦# π₯# β =1 π# π# The graph of the equation π₯π¦ = π, where k is constant, is also a HYPERBOLA. Its center is at the origin, and its ASYMPTOPES are the π₯ and π¦ axes. There are two standard forms for a HYPERBOLA defined by the following characteristics:
Standard Form
(x β h)# (y β k)# β β1 a# b#
(y β k)# (x β h)# β β1 b# a#
(h, k)
(h, k)
Opens left and right
Opens up and down
Center Orientation
h + a, k and (h β a, k)
Vertices Slope of
Β±
Asymptotes Equations of Asymptotes
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b a
b y = k Β± (x β h) a
h, b + k and (h, b β k) Β±
b a
b y = k Β± (x β h) a
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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 straight lines that you may see illustrated along with the Hyperbola, these are called the ASYMPTOTES. The ASYMPTOTES are not officially part of the HYPERBOLA, they help establish the center of the HYPERBOLA which is located at the point where these two lines intersect.
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CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Determine the general quadratic equation of a hyperbola with asymptotes along the x and y axes and which 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: Information revolving around the characteristics 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 takes the standard form π₯π¦ = π. Since the hyperbola passes through the point (3,2), we can take this general formula, plug in our point, and determine what our value of k will be. Doing so we get: π = π₯π¦ = 3 2 = 6
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The general quadratic equation of this hyperbola is found by replacing βπβ in the equation π₯π¦ = π, such that: π₯π¦ = 6 Or: π₯π¦ β 6 = 0 The correct answer choice is C. ππ β π = π CASE π CIRCLES: π = π Information revolving around CIRCLES 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).
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The standard circle can be illustrated as:
The FORMULA of a CIRCLE in STANDARD FORM 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 equation of a CIRCLE centered at the coordinates (β, π) and having a radius βπβ, can be written as:
π=
π₯ββ
#
+ π¦βπ
#
Or: π₯ββ
#
+ π¦βπ
#
= π#
A TANGENT to a circle is a line with an endpoint that doesnβt fall within the CIRCLE and that intersects the circle in one and only one point.
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Every TANGENT to a circle is perpendicular to a radius drawn to the point of tangency. The FORMULA for defining 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 general formula: π‘# = π₯β’ β β
#
+ π¦β’ β π
#
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β π#
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CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
The equation that best represents a circle having a radius of 6 and centered at β3,4 is most close to: 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 of a CIRCLE in STANDARD FORM 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
#
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Or: π₯ββ
#
+ π¦βπ
#
= π#
The problem statement defines the center of our circle, which is: (β, π) = (β3, 4) We are also told that the radius of the circle will be: π=6 Plugging these data points into the standard equation provides the result: π₯+3
#
+ π¦β4
#
= 36
The correct answer choice is A. π + π
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π
+ πβπ
π
= ππ
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CONCEPT INTRO: An equation with second-order terms shown in quadratic form will produce a CONIC SECTION when graphed in a Cartesian coordinate system. The GENERAL FORM of a 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 written as: Ax # + Bxy + Cy # + Dx + Ey + F = 0 Where both π΄ πππ πΆ are real, non-zero numbers. The RULES to help us 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 can quickly classify the type of conic section using the discriminant, such that: 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 written as: 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.
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The ECCENTRICITY of a conic section is defined as:
π = πΈπΆπΆπΈπππ πΌπΆπΌππ =
cos π cos π
CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Classify the shape of the graph given the general quadratic equation: 4π₯ # β 4π₯π¦ + π¦ # β 5 5π₯ + 5 = 0 A. Ellipse B. Hyperbola C. Circle D. Parabola
SOLUTION: The GENERAL FORM of a 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.
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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 Letβs work the formula we are given so that it is presented to us in this general form. In this problem, we are given: 4π₯ # β 4π₯π¦ + π¦ # β 5 5π₯ + 5 = 0 Referring to the general quadratic form of a conic section, it appears everything is already set up as we would want it. With that being established, we find that: A=4 B = -4 C=1 The RULES to help us 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. To classify the type of conic section, we must first calculate the DISCRIMINANT given the data from the formula stated.
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The DISCRIMINANT can be defined using the formula: π΅ # β 4π΄πΆ With all the proper variables define, we simply just need to plug and play, giving us: π΅ # β 4π΄πΆ = β4
#
β4 4 1 =0
With the DISCRIMINANT defined, we refer to the set of RULES stated for us in the NCEES REFERENCE HANDBOOK, which state that: 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. With a DISCRIMANT of zero, we can conclude that the graph of the CONIC SECTION will be the shape of a PARABOLA. The correct answer choice is D. π·πππππππ CASE π: PARABOLAS π = π Information revolving around 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. Made with
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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 can be written as: π π₯ = ππ₯ # + ππ₯ + π When plotted out, the graph of a quadratic function 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. On either side of this line, the parabola is a one to one mirror image of itself. Therefore, if one point on a parabola is known, then the point directly on the other side is known as well, based on the AXIS OF SYMMETRY. The highest or lowest point of a parabola is called the VERTEX. The GENBERAL FORMULA in 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)
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Where: β’ The center of the parabola is located at the coordinates β, π β’ The focus of the parabola is located at runs along the coordinate: π₯ = β
f #
Given this general formula, a parabola can be plotted as such:
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. Therefore, we must take note to retain 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:
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π₯ # = 4ππ¦ Where: β’ If π > 0, the parabola opens UPWARD β’ If π < 0, the parabola opens DOWNWARD. Illustrating this, we have:
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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Generally illustrating this, we have:
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CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
The general quadratic equation of a parabola with a vertex at the origin and focus located at (β2, 0) is best represented as: A. π¦ # + 3π₯ = 0 B. π¦ # + 5π₯ = 0 C. π¦ # + 8π₯ = 0 D. π¦ # + 13π₯ = 0
SOLUTION: Information revolving around 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 a vertex at the origin and a focus on the x-axis takes the general form: π¦ # = 4ππ₯ Where (π, 0) is the coordinates of the focus.
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Since the x-coordinate of the focus is defined as β2, which is given in our problem statement, the general quadratic equation of this parabola can be defined by replacing βπβ with β2 such that: π¦ # = 4(β2)π₯ Which gives us: π¦ # = β8π₯ Rearranging this formula so that it is in standard quadratic form, we find: π¦ # + 8π₯ = 0 The correct answer choice is C. ππ + ππ = π
CASE π: ELLIPSES π < π Information revolving around 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 lays out geometrically as an βOvalβ when presented in your typical x-y coordinate system.
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Every ELLIPSE has two fixed points called the FOCI, wherein which, taking any single point on this ellipse, the sum of the lengths from that point to each of the FOCI remains CONSTANT. This characteristic is what drives the shape of the ELLIPSE. This CONSTANT summation of lengths is also the length of the MAJOR AXIS for an ELLIPSE. An ELLIPSE can be generally illustrated as:
The FORMULA for an ELLIPSE in STANDARD FORM 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 for an ellipse is represented by the equation: π₯ββ π#
#
π¦βπ + π#
#
=1
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Where: β’ The the coordinates β, π represents that center of the ellipse β’ The variable βπβ gives us the distance from the center x-coordinate to the right and left most point of the ELLIPSE β’ The variable βπβ gives us the distance from the center y-coordinate to the upper and lower most point of the ELLIPSE An ELLIPSE can quickly be plotted after defining the four extreme points, such that: 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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Where: β’ βaβ is the distance from the center of the ELLIPSE to the LEFT or RIGHT most point β’ βbβ is the distance from the center of the ELLIPSE to the UPPER or LOWER most point β’ βcβ is the FIXED DISTANCE from the center of the ELLIPSE to either of the FOCI The FORMULAS for locating the 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±
π π
An ELLIPSE that is centered at the origin β, π = (0,0) and has itβs FOCI on the x-axis is represented by the equation: π₯# π¦# + =1 π# π#
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An ELLIPSE that is centered at the origin β, π = (0,0) and has itβs FOCI on the y-axis is represented by the equation: π¦# π₯# + =1 π# π# ELLIPSE TYPE 1: HORIZONTAL ORIENTATION The FORMULA in 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 is represented in standard form as: π₯ββ π#
#
π¦βπ + π#
#
=1
An ELLIPSE with a HORIZONTAL ORIENTATION can generally be drawn as:
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ELLIPSE TYPE 2: VERTICAL ORIENTATION The FORMULA in 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. Typically, we will see that variable βaβ tied directly to the x-term in the general form of an ELLIPSEβ¦resulting in an ellipse with a HORIZONTAL ORIENTATION as described above. However, when βaβ is tied directly to the y-term, then the ORIENTATION of the ELLIPSE flips, presenting itself VERTICALLY. The general form of an ELLIPSE with a VERTICAL ORIENTATION can be written as: π¦ββ π#
#
π₯βπ + π#
#
=1
An ELLIPSE with a VERTICAL ORIENTATION can generally be drawn as:
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CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
The center of the ellipse driven by the defined function is most close to: 4 π₯+1
#
+ π¦β3
#
=1
A. β1, 3 B. 1, 3 C. 1, β 3 D. β1, β3
SOLUTION: The FORMULA of an ELLIPSE in STANDARD FORM 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
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The first step is to rearrange the formula that we are given in to a form that best represents this standard form. Doing so we get: π₯+1 1 4
#
+ π¦β3
#
=1
Note that there isnβt a variable below the y term, which indicates that π # = 1. We can place this value in to the equation so that it replicates the standard form: π₯+1 1 4
#
π¦β3 + 1
#
=1
In the standard form, the point (β, π) is the center of the ellipse. Revisiting where these variables fall within the formula, the standard formula reads: π₯ββ π#
#
π¦βπ + π#
#
=1
Therefore, the center of the ellipse is given by the coordinates (β, π) which are: (β1,3) The correct answer choice is A. βπ, π
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CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
The right most point of the ellipse driven by the defined function is most close to: 4 π₯+1
#
+ π¦β3
#
=1
s
A. (β , 3) #
s
B. ( , β3) #
s
C. (β , β3) #
D.
s #
,3
SOLUTION: The FORMULA of an ELLIPSE in STANDARD FORM 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 can quickly be plotted after defining the four extreme points, such that: β’ Right Most Point: β + π, π β’ Left Most Point: (β β π, π)
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β’ Top Most Point: (β, π + π) β’ Bottom Most Point: (β, π β π) In this problem, we are only concerned with the RIGHT MOST POINT: β’ Right Most Point: β + π, π We wonβt be able to determine this point without definition on where the center of this ellipse falls, lets get this point. 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 formula that we are given in to a form that best represents this standard form. Doing so we get: π₯+1 1 4
#
+ π¦β3
#
=1
Note that there isnβt a variable below the y term, which indicates that π # = 1.
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We can place this value in to the equation so that it replicates the standard form: π₯+1 1 4
#
π¦β3 + 1
#
=1
In the standard form, the point (β, π) is the center of the ellipse. Revisiting where these variables fall within the formula, the standard formula reads: π₯ββ π#
#
π¦βπ + π#
#
=1
Therefore, the center of the ellipse is given by the coordinates β, π which are: (β1,3) We can now define the values of the variable π as it is represented in this formula, we are told that: 1 π# = 4 Which makes: 1 π= 2
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The right most point is given by the coordinates: 1 β + π, π = β1 + , 3 2 Or: 1 β ,3 2 π
The correct answer choice is A. (β , π) π
CASE π HYPERBOLA: π > π Information revolving around 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 lays out geometrically as two reverse facing parabolas in typical x-y coordinate system. Every HYPERBOLA has two fixed points called the FOCI, wherein which, taking any single point on this hyperbola, the difference of the lengths from that point to each of the FOCI remains CONSTANT. This characteristic is what drives the shape of the HYPERBOLA.
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This CONSTANT difference of lengths is also the length of the TRANSVERSE AXIS for the HYPERBOLA. A HYPERBOLA can be generally illustrated as:
The FORMULA for a HYPERBOLA in the STANDARD FORM 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 HYPERBOLA is written as: π₯ββ π#
#
π¦βπ β π#
#
=1
Where: β’ The center of the HYPERBOLA is located at the coordinates β, π
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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 to define 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 a HYPERBOLA is represented by the coordinates: (Β±ππ, 0) The DIRECTRIX of a HYPERBOLA is represented by the equation:
π₯=Β±
π π
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A HYPERBOLA that is centered at the origin β, π = (0,0) and has a HYPERBOLA 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 FOCI on the y-axis, is represented by the equation: π¦# π₯# β =1 π# π# The graph of the equation π₯π¦ = π, where k is constant, is also a HYPERBOLA. Its center is at the origin, and its ASYMPTOPES are the π₯ and π¦ axes. There are two standard forms for a HYPERBOLA defined by the following characteristics:
Standard Form
(x β h)# (y β k)# β β1 a# b#
(y β k)# (x β h)# β β1 b# a#
(h, k)
(h, k)
Opens left and right
Opens up and down
Center Orientation
h + a, k and (h β a, k)
Vertices Slope of
Β±
Asymptotes Equations of Asymptotes
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b a
b y = k Β± (x β h) a
h, b + k and (h, b β k) Β±
b a
b y = k Β± (x β h) a
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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 straight lines that you may see illustrated along with the Hyperbola, these are called the ASYMPTOTES. The ASYMPTOTES are not officially part of the HYPERBOLA, they help establish the center of the HYPERBOLA which is located at the point where these two lines intersect.
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CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Determine the general quadratic equation of a hyperbola with asymptotes along the x and y axes and which 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: Information revolving around the characteristics 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 takes the standard form π₯π¦ = π. Since the hyperbola passes through the point (3,2), we can take this general formula, plug in our point, and determine what our value of k will be. Doing so we get: π = π₯π¦ = 3 2 = 6
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The general quadratic equation of this hyperbola is found by replacing βπβ in the equation π₯π¦ = π, such that: π₯π¦ = 6 Or: π₯π¦ β 6 = 0 The correct answer choice is C. ππ β π = π CASE π CIRCLES: π = π Information revolving around CIRCLES 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).
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The standard circle can be illustrated as:
The FORMULA of a CIRCLE in STANDARD FORM 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 equation of a CIRCLE centered at the coordinates (β, π) and having a radius βπβ, can be written as:
π=
π₯ββ
#
+ π¦βπ
#
Or: π₯ββ
#
+ π¦βπ
#
= π#
A TANGENT to a circle is a line with an endpoint that doesnβt fall within the CIRCLE and that intersects the circle in one and only one point.
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Every TANGENT to a circle is perpendicular to a radius drawn to the point of tangency. The FORMULA for defining 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 general formula: π‘# = π₯β’ β β
#
+ π¦β’ β π
#
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CONIC SECTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
The equation that best represents a circle having a radius of 6 and centered at β3,4 is most close to: 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 of a CIRCLE in STANDARD FORM 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
#
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Or: π₯ββ
#
+ π¦βπ
#
= π#
The problem statement defines the center of our circle, which is: (β, π) = (β3, 4) We are also told that the radius of the circle will be: π=6 Plugging these data points into the standard equation provides the result: π₯+3
#
+ π¦β4
#
= 36
The correct answer choice is A. π + π
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π
+ πβπ
π
= ππ
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