33.1 Conic Sections Problem Set

CONIC SECTIONS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: Determine the center of the ellipse defined by the function: 𝑥2 𝑦2 + =1 100 64 A. (−10, 10) B. (0,0) C. (−5, 5) D. (10,8)

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PROBLEM 2: A circle with its center located at the point (1, −2) and having a radius of 3 is plotted on your standard x-y coordinate system. If a single tangent line intersects the outer edge of this circle and passes through the point (5, −5), the equation of the circle will be best represented as: A. (𝑥 − 1)2 + (𝑦 + 2)2 = 9 B. (𝑥 − 2)2 + (𝑦 + 1)2 = 14 C. (𝑥 − 2)2 + (𝑦 + 7)2 = 19 D. (𝑥 − 5)2 + (𝑦 + 2)2 = 23

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PROBLEM 3: The conic section resulting from the stated equation is best represented as: 4𝑥 2 − 𝑦 2 + 8𝑥 + 4𝑦 + 2𝑥𝑦 = 15 A. Circle B. Ellipse C. Parabola D. Hyperbola

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PROBLEM 4: Determine the center of the ellipse represented by the function: 25𝑥 2 + 𝑦 2 = 25 A. (−10, 10) B. (0, 0) C. (−5, 5) D. (0, 10)

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PROBLEM 5: Determine the center of the ellipse represented by the function: 𝑥 2 (𝑦 − 3)2 + =1 49 4 A. (−10, 10) B. (0, 0) C. (−5, 5) D. (0, 3)

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PROBLEM 6: Determine the center of the circle resulting from the equation: 𝑥 2 − 6𝑥 + 𝑦 2 + 10𝑦 + 25 = 0 A. (1, −2) B. (3, −5) C. (1, −15) D. (2, −9)

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PROBLEM 7: The points of intersection between the circle defined by 𝑥 2 + 𝑦 2 − 𝑥 − 3𝑦 = 0 and the line 𝑦 = 𝑥 − 1 are most close to: A. (−1,0) and (2,1) B. (1,0) and (2,1) C. (2,0) and (2,1) D. (2,2) and (1, 2)

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PROBLEM 8: The center of the circle given by the equation is most close to: 3𝑥 2 + 3𝑦 2 − 12𝑥 + 4 = 0 A. (3, −5) B. (2,0) C. (1, −15) D. (2, −9)

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PROBLEM 9: Classify the graph of the general quadratic equation given by the function: 8𝑦 2 + 6𝑥𝑦 − 9 = 0 A. Ellipse B. Hyperbola C. Circle D. Parabola

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PROBLEM 10: The radius of the circle defined by 𝑥 2 + 𝑦 2 − 4𝑥 + 8𝑦 = 7 is most close to: A. √3 B. 2√5 C. 3√3 D. 4√3

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PROBLEM 11: Determine the right most point of the ellipse represented by the function: 𝑥2 𝑦2 + =1 100 64 A. (5, 0) B. (10, 0) C. (10, 5) D. (5, 5)

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PROBLEM 12: Determine the left most point of the ellipse represented by the function: 𝑥2 𝑦2 + =1 100 64 A. (−5, 0) B. (−10, 0) C. (−10, −5) D. (−10, −5)

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PROBLEM 13: Determine the top most point of the ellipse represented by the function: 𝑥2 𝑦2 + =1 100 64 A. (0, 8) B. (0, 9) C. (0, −8) D. (8, 8)

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PROBLEM 14: Determine the bottom most point of the ellipse represented by the function: 𝑥2 𝑦2 + =1 100 64 A. (−8, 0) B. (0, −9) C. (0, 8) D. (0, −8)

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PROBLEM 15: Determine the right most point of the ellipse represented by the function: 25𝑥 2 + 𝑦 2 = 25 A. (1, 0) B. (0, 0) C. (−5, 5) D. (5, 1)

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PROBLEM 16: Determine the left most point of the ellipse represented by the function: 25𝑥 2 + 𝑦 2 = 25 A. (1, 0) B. (0, 0) C. (−1, 0) D. (5, 1)

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PROBLEM 17: Determine the top most point of the ellipse represented by the function: 25𝑥 2 + 𝑦 2 = 25 A. (1, 0) B. (0, 1) C. (5, 0) D. (0,5)

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PROBLEM 18: Determine the bottom most point of the ellipse represented by the function: 25𝑥 2 + 𝑦 2 = 25 A. (1, 5) B. (−5, 1) C. (5, −5) D. (0, −5)

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PROBLEM 19: Determine the right most point of the ellipse represented by the function: 𝑥 2 (𝑦 − 3)2 + =1 49 4 A. (7,3) B. (3,7) C. (−7, 0) D. (0, 7)

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PROBLEM 20: Determine the left most point of the ellipse represented by the function: 𝑥 2 (𝑦 − 3)2 + =1 49 4 A. (7, 0) B. (0, 3) C. (−7, 3) D. (7, 3)

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PROBLEM 21: Determine the top most point of the ellipse represented by the function: 𝑥 2 (𝑦 − 3)2 + =1 49 4 A. (1, 0) B. (0, 1) C. (5, 0) D. (0,5)

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PROBLEM 22: Determine the bottom most point of the ellipse represented by the function: 𝑥 2 (𝑦 − 3)2 + =1 49 4 A. (1, 5) B. (−5, 1) C. (5, −5) D. (0,1)

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PROBLEM 23: Determine the radius of the circle resulting from the equation: 𝑥 2 − 6𝑥 + 𝑦 2 + 10𝑦 + 25 = 0 A. 2 B. 3 C. 4 D. 5

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PROBLEM 24: The radius of the circle given by the equation is most close to: 3𝑥 2 + 3𝑦 2 − 12𝑥 + 4 = 0

8

A. √3 3

B. √2 C. 4√2 D. 𝑟 = 5

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PROBLEM 25: The center of the circle defined by 𝑥 2 + 𝑦 2 − 4𝑥 + 8𝑦 = 7 is located at: A. (2, −4) B. (5, −2) C. (−2,5) D. (−2, −4)

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CONIC SECTIONS | SOLUTIONS SOLUTION 1: 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. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 𝑥2 𝑦2 + =1 100 64 We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: 𝑥2 𝑦2 + =1 102 82

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We now know that: 𝑎 = 10 𝑏=8 The point (ℎ, 𝑘) will be the center of the ellipse which will be located at (0, 0). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: 𝑥2 𝑦2 + =1 102 82 Which can equivalently written as: (𝑥 − 0)2 (𝑦 − 0)2 + =1 102 82 This confirms our center point being located at: (ℎ, 𝑘) = (0,0) The correct answer choice is B. (0,0) Made with

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SOLUTION 2: 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. The STANDARD FORM OF A CIRCLE centered at coordinates (ℎ, 𝑘) with radius “𝑟” is represented by the equation:

𝑟 = √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 Or: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 In this problem we are given: Center Coordinates: (1, −2) Radius: 3 Tangent Line Point 1: (5, −5) Which means that we have everything we need to define the equation of the circle. But what about this whole TANGENT, what are we going to do with that information? Ignore it. Made with

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A lot of times you are going to be presented with information that is irrelevant to the problem you are working. The key is to know the concepts in which you are being questioned on and vet all the information according to that. In this case, we are only concerned with the equation of a circle…the problem statement gives us everything we need to define it. The equation of a circle with radius 3 and a center at the origin (0,0) is given as: 𝑥2 + 𝑦2 = 9 By translation of axis, the equation of this circle can be moved so that it has it’s center located at (1, −2), doing so we get: (𝑥 − 1)2 + (𝑦 + 2)2 = 9

The correct answer choice is A. (𝒙 − 𝟏)𝟐 + (𝒚 + 𝟐)𝟐 = 𝟗

SOLUTION 3: 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 written as: Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 Where both 𝐴 𝑎𝑛𝑑 𝐶 are real, non-zero numbers. In the problem statement, we are given the quadratic expression: 4𝑥 2 − 𝑦 2 + 8𝑥 + 4𝑦 + 2𝑥𝑦 = 15 Referring back to our general quadratic form of a conic section, we have: 𝐴𝑥 2 + 𝐵𝑥𝑦 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 Which tells us that: 𝐴=4 𝐵=2 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. We can quickly classify the type of conic section using the DISCRIMINANT, such that: • If 𝐵 2 − 4𝐴𝐶 < 0, the conci section is an ELLIPSE Made with

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• If 𝐵 2 − 4𝐴𝐶 > 0, the conic section is a HYPERBOLA. • If 𝐵 2 − 4𝐴𝐶 = 0, the conic section is a PARABOLA • If 𝐴 = 𝐶 and 𝐵 = 0, the conic section is a CIRCLE. • If 𝐴 = 𝐵 = 𝐶 = 0, the function defines a STRAIGHT LINE. Calculating the DISCRIMINANT gives us: 𝐵 2 − 4𝐴𝐶 = (2)2 − 4(4)(−1) = 20 Referring back to our set of RULES, we see that since 𝐵 2 − 4𝐴𝐶 > 0, the conic section resulting from this equation is a HYPERBOLA. The correct answer choice is D. 𝑯𝒚𝒑𝒆𝒓𝒃𝒐𝒍𝒂

SOLUTION 4: 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. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2

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The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 25𝑥 2 + 𝑦 2 = 25 We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: 𝑦2 𝑥 + =1 25 2

Or taking it one more step: 𝑥2 𝑦2 + =1 12 52 We now know that: 𝑎=1 𝑏=5 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 0). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Made with

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Our formula reads: 𝑥2 𝑦2 + =1 12 52 Which can equivalently written as: (𝑥 − 0)2 (𝑦 − 0)2 + =1 12 52 This confirms our center point being located at: (ℎ, 𝑘) = (0,0) The correct answer choice is B. (0,0)

SOLUTION 5: 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. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2

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The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 𝑥 2 (𝑦 − 3)2 + =1 49 4 The NUMERATORS look fairly good, let’s add our “h” value just to make it match even more: (𝑥 − 0)2 (𝑦 − 3)2 + =1 49 4 We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: (𝑥 − 0)2 (𝑦 − 3)2 + =1 72 22 With this formula now fully in line with our general formula of an ELLIPSE, we now know that: 𝑎=7 𝑏=2 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 3). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. Made with

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(𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: (𝑥 − 0)2 (𝑦 − 3)2 + =1 72 22 This confirms our center point being located at: (ℎ, 𝑘) = (0,3) The correct answer choice is D. (0,3)

SOLUTION 6: 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. The STANDARD FORM OF A CIRCLE centered at coordinates (ℎ, 𝑘) with radius “𝑟” is represented by the equation:

𝑟 = √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2

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Or: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 In this problem we are given the equation: 𝑥 2 − 6𝑥 + 𝑦 2 + 10𝑦 + 25 = 0 The first thing we need to do is break this formula down so that it is more representative of our standard form. We can factor the original equation such that: (𝑥 − 3)(𝑥 + 3) + (𝑦 + 5)(𝑦 + 5) = 0 We now that we need to get both the x and y terms in the form: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 We can do that with the y term as it stands: (𝑥 − 3)(𝑥 + 3) + (𝑦 + 5)2 = 0 But the x term still needs to be sorted out.

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Not a problem, let’s make both the x factors (𝑥 − 3), such that: (𝑥 − 3)2 − 9 + (𝑦 + 5)2 = 0 We can do that because we are also subtracting 9 out on the left side, which is the result of changing the sign in the one factor. Throwing our constant variable on the right side of the equation, we get: (𝑥 − 3)2 + (𝑦 + 5)2 = 9 With this equation now in the standard form, let’s refer back to our standard form of a circle: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 This tells us that the center (ℎ, 𝑘) of this circle is located at: Center Coordinates: (3, −5) The correct answer choice is B. (𝟑, −𝟓)

SOLUTION 7: 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. Made with

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The STANDARD FORM OF A CIRCLE centered at coordinates (ℎ, 𝑘) with radius “𝑟” is represented by the equation:

𝑟 = √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 Or: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 In this problem we are given the equation of a circle as: 𝑥 2 + 𝑦 2 − 𝑥 − 3𝑦 = 0 And the line: 𝑦 =𝑥−1 Let’s solve the equations simultaneously by substituting the expression 𝑦 = 𝑥 − 1 into the equation of the circle such that: 𝑥 2 + (𝑥 − 1)2 − 𝑥 − 3(𝑥 − 1) = 0 Multiplying and expanding, we get: 𝑥 2 + 𝑥 2 − 2𝑥 + 1 − 𝑥 − 3𝑥 + 3 = 0

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Combing the like terms we get: 2𝑥 2 − 6𝑥 + 4 = 0 And simplifying: 𝑥 2 − 3𝑥 + 2 = 0 We are not at a point where we can factor, doing so we get: (𝑥 − 1)(𝑥 − 2) = 0 This shows that the x values for the solutions where the equations will intersect are: 𝑥=1 𝑥=2 Plugging these values into the equation 𝑦 = 𝑥 − 1 to find the corresponding y coordinates of these two intersecting points, we get: 𝑦=0 𝑦=1 Therefore, the points of intersection between the circle defined by 𝑥 2 + 𝑦 2 − 𝑥 − 3𝑦 = 0 and the line 𝑦 = 𝑥 − 1 are: (1,0) Made with

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(2,1) The correct answer choice is B. (𝟏, 𝟎) 𝐚𝐧𝐝 (𝟐, 𝟏)

SOLUTION 8: 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. The STANDARD FORM OF A CIRCLE centered at coordinates (ℎ, 𝑘) with radius “𝑟” is represented by the equation:

𝑟 = √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 Or: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 In this problem we are given the equation: 3𝑥 2 + 3𝑦 2 − 12𝑥 + 4 = 0 The first thing we need to do is break this formula down so that it is more representative of our standard form.

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Let’s collect the x and y terms such that: 3𝑥 2 − 12𝑥 + 3𝑦 2 + 4 = 0 Dividing all these terms by 3 we get: 4

𝑥 2 − 4𝑥 + 𝑦 2 + 3 = 0 Moving the constant to the right side of the equation and completing the square on the x-term we get: 4

(𝑥 2 − 4𝑥 + 4) + 𝑦 2 = − + 4 3 We can now factor and simplify to get: 8

(𝑥 − 2)2 + (𝑦 + 0)2 = 3 With this equation now in the standard form, let’s refer back to our standard form of a circle: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 This tells us that the center (ℎ, 𝑘) of this circle is located at: Center Coordinates: (2, 0)

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The correct answer choice is B. (𝟐, 𝟎)

SOLUTION 9: 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 2 + Bxy + Cy 2 + Dx + Ey + F = 0 Where both 𝐴 𝑎𝑛𝑑 𝐶 are real, non-zero numbers. In the problem statement, we are given the quadratic expression: 8𝑦 2 + 6𝑥𝑦 − 9 = 0 Referring back to our general quadratic form of a conic section, we have: 𝐴𝑥 2 + 𝐵𝑥𝑦 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 Which tells us that: 𝐴=0 𝐵=6 𝐶=8 Made with

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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. We can quickly classify the type of conic section using the DISCRIMINANT, such that: • If 𝐵 2 − 4𝐴𝐶 < 0, the conci section is an ELLIPSE • If 𝐵 2 − 4𝐴𝐶 > 0, the conic section is a HYPERBOLA. • If 𝐵 2 − 4𝐴𝐶 = 0, the conic section is a PARABOLA • If 𝐴 = 𝐶 and 𝐵 = 0, the conic section is a CIRCLE. • If 𝐴 = 𝐵 = 𝐶 = 0, the function defines a STRAIGHT LINE. Calculating the DISCRIMINANT gives us: 𝐵 2 − 4𝐴𝐶 = (6)2 − 4(0)(8) = 36 Referring back to our set of RULES, we see that since 𝐵 2 − 4𝐴𝐶 > 0, the conic section resulting from this equation is a HYPERBOLA. The correct answer choice is B. 𝑯𝒚𝒑𝒆𝒓𝒃𝒐𝒍𝒂

SOLUTION 10: 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. Made with

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The STANDARD FORM OF A CIRCLE centered at coordinates (ℎ, 𝑘) with radius “𝑟” is represented by the equation:

𝑟 = √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 Or: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 In this problem we are given the equation: 𝑥 2 + 𝑦 2 − 4𝑥 + 8𝑦 = 7 The first thing we need to do is break this formula down so that it is more representative of our standard form. Let’s start off with grouping the x and y terms and completing the square: 𝑥 2 − 4𝑥 + 4 + 𝑦 2 + 8𝑦 + 16 = 7 And because we added the constant 4 and 16 on the left side, we also have to add them to the right side of the equation: 𝑥 2 − 4𝑥 + 4 + 𝑦 2 + 8𝑦 + 16 = 7 + 4 + 16

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Or: 𝑥 2 − 4𝑥 + 4 + 𝑦 2 + 8𝑦 + 16 = 7 + 4 + 16 Factoring we get: (𝑥 − 2)2 + (𝑦 + 4)2 = 27 With this equation now in the standard form, let’s refer back to our standard form of a circle: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 This tells us that the center (ℎ, 𝑘) of this circle is located at: This tells us that the radius of the circle is:

𝑟 = 3√3 The correct answer choice is C. 𝟑√𝟑

SOLUTION 11: 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. Made with

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The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) • Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the right most point, defined as: • Right Most Point: (ℎ + 𝑎, 𝑘) There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 𝑥2 𝑦2 + =1 100 64

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We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: 𝑥2 𝑦2 + =1 102 82 We now know that: 𝑎 = 10 𝑏=8 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 0). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: 𝑥2 𝑦2 + =1 102 82 Which can equivalently written as: (𝑥 − 0)2 (𝑦 − 0)2 + =1 102 82 Made with

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This confirms our center point being located at: (ℎ, 𝑘) = (0,0) Again the Right Most Point of this ellipse can be defined as: (ℎ + 𝑎, 𝑘) With all of our data defined as: 𝑎 = 10 𝑏=8 (ℎ, 𝑘) = (0,0) Plugging in this data, we find that the Right Most Point is located at: (0 + 10,0) = (10,0) The correct answer choice is B. (𝟏𝟎, 𝟎)

SOLUTION 12: 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.

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The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) • Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the left most point, defined as: • Left Most Point: (ℎ − 𝑎, 𝑘) There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 𝑥2 𝑦2 + =1 100 64

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We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: 𝑥2 𝑦2 + =1 102 82 We now know that: 𝑎 = 10 𝑏=8 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 0). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: 𝑥2 𝑦2 + =1 102 82 Which can equivalently written as: (𝑥 − 0)2 (𝑦 − 0)2 + =1 102 82 Made with

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This confirms our center point being located at: (ℎ, 𝑘) = (0,0) Again the Left Most Point of this ellipse can be defined as: (ℎ − 𝑎, 𝑘) With all of our data defined as: 𝑎 = 10 𝑏=8 (ℎ, 𝑘) = (0,0) Plugging in this data, we find that the Left Most Point is located at: (0 − 10,0) = (−10,0) The correct answer choice is B. (-10, 0)

SOLUTION 13: 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.

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The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) • Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the top most point, defined as: • Top Most Point: (ℎ, 𝑘 + 𝑏) There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 𝑥2 𝑦2 + =1 100 64

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We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: 𝑥2 𝑦2 + =1 102 82 We now know that: 𝑎 = 10 𝑏=8 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 0). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: 𝑥2 𝑦2 + =1 102 82 Which can equivalently written as: (𝑥 − 0)2 (𝑦 − 0)2 + =1 102 82 Made with

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This confirms our center point being located at: (ℎ, 𝑘) = (0,0) Again the Top Most Point of this ellipse can be defined as: (ℎ, 𝑘 + 𝑏) With all of our data defined as: 𝑎 = 10 𝑏=8 (ℎ, 𝑘) = (0,0) Plugging in this data, we find that the Top Most Point is located at: (0,0 + 8) = (0,8) The correct answer choice is A. (0,8)

SOLUTION 14: 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.

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The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) • Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the bottom most point, defined as: • Top Most Point: (ℎ, 𝑘 − 𝑏) There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 𝑥2 𝑦2 + =1 100 64

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We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: 𝑥2 𝑦2 + =1 102 82 We now know that: 𝑎 = 10 𝑏=8 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 0). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: 𝑥2 𝑦2 + =1 102 82 Which can equivalently written as: (𝑥 − 0)2 (𝑦 − 0)2 + =1 102 82 Made with

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This confirms our center point being located at: (ℎ, 𝑘) = (0,0) Again the Bottom Most Point of this ellipse can be defined as: (ℎ, 𝑘 − 𝑏) With all of our data defined as: 𝑎 = 10 𝑏=8 (ℎ, 𝑘) = (0,0) Plugging in this data, we find that the Bottom Most Point is located at: (0,0 − 8) = (0, −8) The correct answer choice is D. (0,-8)

SOLUTION 15: 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.

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The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) • Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the right most point, defined as: • Right Most Point: (ℎ + 𝑎, 𝑘) There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 25𝑥 2 + 𝑦 2 = 25

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We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: 𝑦2 𝑥 + =1 25 2

Or taking it one more step: 𝑥2 𝑦2 + =1 12 52 We now know that: 𝑎=1 𝑏=5 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 0). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: 𝑥2 𝑦2 + =1 12 52 Made with

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Which can equivalently written as: (𝑥 − 0)2 (𝑦 − 0)2 + =1 12 52 This confirms our center point being located at: (ℎ, 𝑘) = (0,0) Again the Right Most Point of this ellipse can be defined as: (ℎ + 𝑎, 𝑘) With all of our data defined as: 𝑎=1 𝑏=5 (ℎ, 𝑘) = (0,0) Plugging in this data, we find that the Right Most Point is located at: (0 + 1,0) = (1,0) The correct answer choice is A. (1,0)

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SOLUTION 16: 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. The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) • Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the left most point, defined as: • Left Most Point: (ℎ − 𝑎, 𝑘) There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Made with

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The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 25𝑥 2 + 𝑦 2 = 25 We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: 𝑦2 𝑥 + =1 25 2

Or taking it one more step: 𝑥2 𝑦2 + =1 12 52 We now know that: 𝑎=1 𝑏=5 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 0). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Made with

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Our formula reads: 𝑥2 𝑦2 + =1 12 52 Which can equivalently written as: (𝑥 − 0)2 (𝑦 − 0)2 + =1 12 52 This confirms our center point being located at: (ℎ, 𝑘) = (0,0) Again the Left Most Point of this ellipse can be defined as: (ℎ − 𝑎, 𝑘) With all of our data defined as: 𝑎=1 𝑏=5 (ℎ, 𝑘) = (0,0) Plugging in this data, we find that the Left Most Point is located at: (0 − 1,0) = (−1,0) Made with

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The correct answer choice is C. (-1,0)

SOLUTION 17: 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. The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) • Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the top most point, defined as: • Top Most Point: (ℎ, 𝑘 + 𝑏) There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse.

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The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 25𝑥 2 + 𝑦 2 = 25 We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that:

𝑥2 +

𝑦2 =1 25

Or taking it one more step: 𝑥2 𝑦2 + =1 12 52 We now know that: 𝑎=1 𝑏=5 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 0). Made with

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We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: 𝑥2 𝑦2 + =1 12 52 Which can equivalently written as: (𝑥 − 0)2 (𝑦 − 0)2 + =1 12 52 This confirms our center point being located at: (ℎ, 𝑘) = (0,0) Again the Top Most Point of this ellipse can be defined as: (ℎ, 𝑘 + 𝑏) With all of our data defined as: 𝑎=1 𝑏=5 Made with

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(ℎ, 𝑘) = (0,0) Plugging in this data, we find that the Top Most Point is located at: (0,0 + 5) = (0,5) The correct answer choice is D. (0,5)

SOLUTION 18: 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. The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) • Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the bottom most point, defined as: • Bottom Most Point: (ℎ, 𝑘 − 𝑏)

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There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 25𝑥 2 + 𝑦 2 = 25 We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: 𝑦2 𝑥 + =1 25 2

Or taking it one more step: 𝑥2 𝑦2 + =1 12 52

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We now know that: 𝑎=1 𝑏=5 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 0). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: 𝑥2 𝑦2 + =1 12 52 Which can equivalently written as: (𝑥 − 0)2 (𝑦 − 0)2 + =1 12 52 This confirms our center point being located at: (ℎ, 𝑘) = (0,0)

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Again the Bottom Most Point of this ellipse can be defined as: (ℎ, 𝑘 − 𝑏) With all of our data defined as: 𝑎=1 𝑏=5 (ℎ, 𝑘) = (0,0) Plugging in this data, we find that the Bottom Most Point is located at: (0,0 − 5) = (0, −5) The correct answer choice is D. (0,-5)

SOLUTION 19: 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. The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) Made with

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• Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the right most point, defined as: • Right Most Point: (ℎ + 𝑎, 𝑘) There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 𝑥 2 (𝑦 − 3)2 + =1 49 4 The NUMERATORS look fairly good, let’s add our “h” value just to make it match even more: (𝑥 − 0)2 (𝑦 − 3)2 + =1 49 4 Made with

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We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: (𝑥 − 0)2 (𝑦 − 3)2 + =1 72 22 With this formula now fully in line with our general formula of an ELLIPSE, we now know that: 𝑎=7 𝑏=2 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 3). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: (𝑥 − 0)2 (𝑦 − 3)2 + =1 72 22 This confirms our center point being located at: (ℎ, 𝑘) = (0,3) Made with

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Again the Right Most Point of this ellipse can be defined as: (ℎ + 𝑎, 𝑘) With all of our data defined as: 𝑎=7 𝑏=2 (ℎ, 𝑘) = (0,3) Plugging in this data, we find that the Right Most Point is located at: (0 + 7,3) = (7,3) The correct answer choice is A. (7,3)

SOLUTION 20: 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. The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) Made with

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• Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the left most point, defined as: • Left Most Point: (ℎ − 𝑎, 𝑘) There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 𝑥 2 (𝑦 − 3)2 + =1 49 4 The NUMERATORS look fairly good, let’s add our “h” value just to make it match even more: (𝑥 − 0)2 (𝑦 − 3)2 + =1 49 4 Made with

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We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: (𝑥 − 0)2 (𝑦 − 3)2 + =1 72 22 With this formula now fully in line with our general formula of an ELLIPSE, we now know that: 𝑎=7 𝑏=2 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 3). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: (𝑥 − 0)2 (𝑦 − 3)2 + =1 72 22 This confirms our center point being located at: (ℎ, 𝑘) = (0,3) Made with

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Again the Left Most Point of this ellipse can be defined as: (ℎ − 𝑎, 𝑘) With all of our data defined as: 𝑎=7 𝑏=2 (ℎ, 𝑘) = (0,3) Plugging in this data, we find that the Left Most Point is located at: (0 − 7,3) = (−7,3) The correct answer choice is C. (-7,3)

SOLUTION 21: 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. The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) Made with

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• Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the top most point, defined as: • Top Most Point: (ℎ, 𝑘 + 𝑏) There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 𝑥 2 (𝑦 − 3)2 + =1 49 4 The NUMERATORS look fairly good, let’s add our “h” value just to make it match even more: (𝑥 − 0)2 (𝑦 − 3)2 + =1 49 4 Made with

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We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: (𝑥 − 0)2 (𝑦 − 3)2 + =1 72 22 With this formula now fully in line with our general formula of an ELLIPSE, we now know that: 𝑎=7 𝑏=2 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 3). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: (𝑥 − 0)2 (𝑦 − 3)2 + =1 72 22 This confirms our center point being located at: (ℎ, 𝑘) = (0,3) Made with

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Again the Top Most Point of this ellipse can be defined as: (ℎ, 𝑘 + 𝑏) With all of our data defined as: 𝑎=7 𝑏=2 (ℎ, 𝑘) = (0,3) Plugging in this data, we find that the Top Most Point is located at: (0,3 + 2) = (0,5) The correct answer choice is D. (0,5)

SOLUTION 22: 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. The four extrema of an ellipse are calculated using the following relationships: • Right Most Point: (ℎ + 𝑎, 𝑘) • Left Most Point: (ℎ − 𝑎, 𝑘) Made with

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• Top Most Point: (ℎ, 𝑘 + 𝑏) • Bottom Most Point: (ℎ, 𝑘 − 𝑏) In this problem, we are concerned only with the bottom most point, defined as: • Bottom Most Point: (ℎ, 𝑘 − 𝑏) There are a few pieces of information that we must define first before we can determine this point. First is the center of the ellipse. The standard form of the equation for an ellipse can be written as: (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 The problem statement defines specifically the driving function of the ellipse we are evaluating, which is: 𝑥 2 (𝑦 − 3)2 + =1 49 4 The NUMERATORS look fairly good, let’s add our “h” value just to make it match even more: (𝑥 − 0)2 (𝑦 − 3)2 + =1 49 4 Made with

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We can slightly tweak and rewrite this function to reflect the variables in the denominator of each term, such that: (𝑥 − 0)2 (𝑦 − 3)2 + =1 72 22 With this formula now fully in line with our general formula of an ELLIPSE, we now know that: 𝑎=7 𝑏=2 The point (ℎ, 𝑘) will be the center of the ellipse, which will be located at (0, 3). We can confirm this by referring back to the general formula of an ellipse, noting where the variables h and k fall. (𝑥 − ℎ)2 (𝑦 − 𝑘)2 + =1 𝑎2 𝑏2 Our formula reads: (𝑥 − 0)2 (𝑦 − 3)2 + =1 72 22 This confirms our center point being located at: (ℎ, 𝑘) = (0,3) Made with

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Again the Bottom Most Point of this ellipse can be defined as: (ℎ, 𝑘 − 𝑏) With all of our data defined as: 𝑎=7 𝑏=2 (ℎ, 𝑘) = (0,3) Plugging in this data, we find that the Bottom Most Point is located at: (0,3 − 2) = (0,1) The correct answer choice is D. (0,1)

SOLUTION 23: 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. The STANDARD FORM OF A CIRCLE centered at coordinates (ℎ, 𝑘) with radius “𝑟” is represented by the equation:

𝑟 = √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 Made with

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Or: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 In this problem we are given the equation: 𝑥 2 − 6𝑥 + 𝑦 2 + 10𝑦 + 25 = 0 The first thing we need to do is break this formula down so that it is more representative of our standard form. We can factor the original equation such that: (𝑥 − 3)(𝑥 + 3) + (𝑦 + 5)(𝑦 + 5) = 0 We now that we need to get both the x and y terms in the form: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 We can do that with the y term as it stands: (𝑥 − 3)(𝑥 + 3) + (𝑦 + 5)2 = 0 But the x term still needs to be sorted out.

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Not a problem, let’s make both the x factors (𝑥 − 3), such that: (𝑥 − 3)2 − 9 + (𝑦 + 5)2 = 0 We can do that because we are also subtracting 9 out on the left side, which is the result of changing the sign in the one factor. Throwing our constant variable on the right side of the equation, we get: (𝑥 − 3)2 + (𝑦 + 5)2 = 9 With this equation now in the standard form, let’s refer back to our standard form of a circle: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 This tells us that the radius “𝑟” is: Radius: 3 The correct answer choice is B. 𝟑

SOLUTION 24: 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. Made with

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The STANDARD FORM OF A CIRCLE centered at coordinates (ℎ, 𝑘) with radius “𝑟” is represented by the equation:

𝑟 = √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 Or: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 In this problem we are given the equation: 3𝑥 2 + 3𝑦 2 − 12𝑥 + 4 = 0, The first thing we need to do is break this formula down so that it is more representative of our standard form. Let’s collect the x and y terms such that: 3𝑥 2 − 12𝑥 + 3𝑦 2 + 4 = 0 Dividing all these terms by 3 we get: 4

𝑥 2 − 4𝑥 + 𝑦 2 + 3 = 0

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Moving the constant to the right side of the equation and completing the square on the x-term we get: 4

(𝑥 2 − 4𝑥 + 4) + 𝑦 2 = − + 4 3 We can now factor and simplify to get: 8

(𝑥 − 2)2 + (𝑦 + 0)2 = 3 With this equation now in the standard form, let’s refer back to our standard form of a circle: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 This tells us that the radius of this circle is:

8

Radius: √3

𝟖

The correct answer choice is A. √𝟑

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SOLUTION 25: 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. The STANDARD FORM OF A CIRCLE centered at coordinates (ℎ, 𝑘) with radius “𝑟” is represented by the equation:

𝑟 = √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 Or: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 In this problem we are given the equation: 𝑥 2 + 𝑦 2 − 4𝑥 + 8𝑦 = 7 The first thing we need to do is break this formula down so that it is more representative of our standard form. Let’s start off with grouping the x and y terms and completing the square: 𝑥 2 − 4𝑥 + 4 + 𝑦 2 + 8𝑦 + 16 = 7

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And because we added the constant 4 and 16 on the left side, we also have to add them to the right side of the equation: 𝑥 2 − 4𝑥 + 4 + 𝑦 2 + 8𝑦 + 16 = 7 + 4 + 16 Or: 𝑥 2 − 4𝑥 + 4 + 𝑦 2 + 8𝑦 + 16 = 7 + 4 + 16 Factoring we get: (𝑥 − 2)2 + (𝑦 + 4)2 = 27 With this equation now in the standard form, let’s refer back to our standard form of a circle: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 This tells us that the center (ℎ, 𝑘) of this circle is located at: (2, −4) The correct answer choice is A. (𝟐, −𝟒)

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