16.5 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 as well as the right most, left most, top most and bottom most points for the following function: 𝑥" 𝑦" + =1 100 64 A. 𝐶𝑒𝑛𝑡𝑒𝑟 −10, 10 , 𝑅𝑖𝑔ℎ𝑡 5, 0 , 𝐿𝑒𝑓𝑡 −5, 0 , 𝑇𝑜𝑝 0, 8 , 𝐵𝑜𝑡𝑡𝑜𝑚 (0, −8) B. 𝐶𝑒𝑛𝑡𝑒𝑟 0, 0 , 𝑅𝑖𝑔ℎ𝑡 10, 0 , 𝐿𝑒𝑓𝑡 −10, 0 , 𝑇𝑜𝑝 0, 9 , 𝐵𝑜𝑡𝑡𝑜𝑚 (0, −9) C. 𝐶𝑒𝑛𝑡𝑒𝑟 −5, 5 , 𝑅𝑖𝑔ℎ𝑡 10, 5 , 𝐿𝑒𝑓𝑡 −10, −5 , 𝑇𝑜𝑝 0, 8 , 𝐵𝑜𝑡𝑡𝑜𝑚 (0, −8) D. 𝐶𝑒𝑛𝑡𝑒𝑟 0, 0 , 𝑅𝑖𝑔ℎ𝑡 10, 0 , 𝐿𝑒𝑓𝑡 −10, 0 , 𝑇𝑜𝑝 0, 8 , 𝐵𝑜𝑡𝑡𝑜𝑚 (0, −8)

SOLUTION 1: 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: 𝑥−ℎ 𝑎"

"

𝑦−𝑘 + 𝑏"

"

=1

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The function of the ellipse we are evaluating is given as: 𝑥" 𝑦" + =1 100 64 We can than rewrite the function to reflect the squared terms in the denominator for 𝑎 𝑎𝑛𝑑 𝑏: 𝑥" 𝑦" + =1 10" 8" We now know that the values are 𝑎 = 10 and 𝑏 = 8 The point (ℎ, 𝑘) is the center of the ellipse, which is (0, 0). The four extrema of an ellipse are calculated as: • Right Most Point: ℎ + 𝑎, 𝑘 = (10,0) • Left Most Point: ℎ − 𝑎, 𝑘 = (−10, 0) • Top Most Point: ℎ, 𝑘 + 𝑏 = (0,8) • Bottom Most Point: ℎ, 𝑘 − 𝑏 = (0, −8)

Therefore, the correct answer choice is D. 𝑪𝒆𝒏𝒕𝒆𝒓 𝟎, 𝟎 , 𝑹𝒊𝒈𝒉𝒕 𝟏𝟎, 𝟎 , 𝑳𝒆𝒇𝒕 −𝟏𝟎, 𝟎 , 𝑻𝒐𝒑 𝟎, 𝟖 , 𝑩𝒐𝒕𝒕𝒐𝒎 (𝟎, −𝟖)



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PROBLEM 2: A tangent to the circle with center (1, −2) and radius 3 passes through point (5, −5). What is the equation of the circle? A. 𝑥 − 1

"

+ 𝑦+2

"

=9

B. 𝑥 − 2

"

+ 𝑦+1

"

= 14

C. 𝑥 − 2

"

+ 𝑦+7

"

= 19

D. 𝑥 − 5

"

+ 𝑦+2

"

= 23

SOLUTION 2: 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 (ℎ, 𝑘) with radius “𝑟” is represented by the equation:

𝑟=

𝑥−ℎ

"

+ 𝑦−𝑘

"

or

𝑥−ℎ

"

+ 𝑦−𝑘

"

= 𝑟"

The equation of a circle with radius 3 and center at the origin is given as: 𝑥" + 𝑦" = 9 By translation of axis, the equation of a circle with center (1, −2) and radius 3 is: 𝑥−1

"

+ 𝑦+2

"

=9

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Therefore, the correct answer choice is A. (𝒙 − 𝟏)𝟐 + (𝒚 + 𝟐)𝟐 = 𝟗 PROBLEM 3: What conic section is described by the following equation? 4𝑥 " − 𝑦 " + 8𝑥 + 4𝑦 + 2𝑥𝑦 = 15 A. 𝐶𝑖𝑟𝑐𝑙𝑒 B. 𝐸𝑙𝑙𝑖𝑝𝑠𝑒 C. 𝑃𝑎𝑟𝑎𝑏𝑜𝑙𝑎 D. 𝐻𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑎

SOLUTION 3: The TOPIC of 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. We are given the quadratic expression: 4𝑥 " − 𝑦 " + 8𝑥 + 4𝑦 + 2𝑥𝑦 = 15 The general quadratic form of a conic section is given by the standard equation: 𝐴𝑥 " + 𝐵𝑥𝑦 + 𝐶𝑦 " + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0



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Where both 𝐴 𝑎𝑛𝑑 𝐶 are not zero. In the quadratic expression provided we find: • 𝐴=4 • 𝐵=2 • 𝐶 = −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. We can define the type of conic section by using the discriminant to classify the properties of the function: • If 𝐵 " − 4𝐴𝐶 < 0, the conci section is an ELLIPSE • If 𝐵 " − 4𝐴𝐶 > 0, the conic section is a HYPERBOLA. • If 𝐵 " − 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. We calculate the discriminant to find: 𝐵 " − 𝐴𝐶 = 2

"

− 4 4 −1 = 20

Since 𝐵 " − 4𝐴𝐶 > 0, the conic section is a 𝐻𝑌𝑃𝐸𝑅𝐵𝑂𝐿𝐴.

Therefore, the correct answer choice is D. 𝑯𝒚𝒑𝒆𝒓𝒃𝒐𝒍𝒂

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PROBLEM 4: Determine the center of the ellipse as well as the right most, left most, top most and bottom most points for the following function: 27𝑥 " + 𝑦 " = 27 A. 𝐶𝑒𝑛𝑡𝑒𝑟 −10, 10 , 𝑅𝑖𝑔ℎ𝑡 5, 0 , 𝐿𝑒𝑓𝑡 −5, 0 , 𝑇𝑜𝑝 0, 8 , 𝐵𝑜𝑡𝑡𝑜𝑚 (0, −8) B. 𝐶𝑒𝑛𝑡𝑒𝑟 0, 0 , 𝑅𝑖𝑔ℎ𝑡 1, 0 , 𝐿𝑒𝑓𝑡 −1, 0 , 𝑇𝑜𝑝 0, 5 , 𝐵𝑜𝑡𝑡𝑜𝑚 (0, −5) C. 𝐶𝑒𝑛𝑡𝑒𝑟 −5, 5 , 𝑅𝑖𝑔ℎ𝑡 10, 5 , 𝐿𝑒𝑓𝑡 −10, −5 , 𝑇𝑜𝑝 0, 8 , 𝐵𝑜𝑡𝑡𝑜𝑚 (0, −8) D. 𝐶𝑒𝑛𝑡𝑒𝑟 0, 10 , 𝑅𝑖𝑔ℎ𝑡 10, 0 , 𝐿𝑒𝑓𝑡 −10, 0 , 𝑇𝑜𝑝 0, 8 , 𝐵𝑜𝑡𝑡𝑜𝑚 (0, −8)

SOLUTION 4: 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: 𝑥−ℎ 𝑎"

"

𝑦−𝑘 + 𝑏"

"

=1

The function of the ellipse we are evaluating is given as: 27𝑥 " + 𝑦 " = 27



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Rearranging the equation, we get: 𝑦" 𝑥 + =1 27 "

We can than rewrite the function to reflect the squared terms in the denominator for 𝑎 𝑎𝑛𝑑 𝑏: 𝑥" 𝑦" + =1 1" 5" We now know that the values are 𝑎 = 1 and 𝑏 = 5 The point (ℎ, 𝑘) is the center of the ellipse, which is (0,0). The four extrema of an ellipse are calculated as: • Right Most Point: ℎ + 𝑎, 𝑘 = (1,0) • Left Most Point: ℎ − 𝑎, 𝑘 = (−1,0) • Top Most Point: ℎ, 𝑘 + 𝑏 = (0,5) • Bottom Most Point: ℎ, 𝑘 − 𝑏 = (0, −5)

Therefore, the correct answer choice is B. 𝑪𝒆𝒏𝒕𝒆𝒓 𝟎, 𝟎 , 𝑹𝒊𝒈𝒉𝒕 𝟏, 𝟎 , 𝑳𝒆𝒇𝒕 −𝟏, 𝟎 , 𝑻𝒐𝒑 𝟎, 𝟓 , 𝑩𝒐𝒕𝒕𝒐𝒎 (𝟎, −𝟓)



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PROBLEM 5: Determine the center of the ellipse as well as the right most, left most, top most and bottom most points for the following function: 𝑥" 𝑦−3 + 49 4

"

=1

A. 𝐶𝑒𝑛𝑡𝑒𝑟 −10, 10 , 𝑅𝑖𝑔ℎ𝑡 5, 0 , 𝐿𝑒𝑓𝑡 −5, 0 , 𝑇𝑜𝑝 0, 8 , 𝐵𝑜𝑡𝑡𝑜𝑚 (0, −8) B. 𝐶𝑒𝑛𝑡𝑒𝑟 0, 0 , 𝑅𝑖𝑔ℎ𝑡 1, 0 , 𝐿𝑒𝑓𝑡 −1, 0 , 𝑇𝑜𝑝 0, 5 , 𝐵𝑜𝑡𝑡𝑜𝑚 (0, −5) C. 𝐶𝑒𝑛𝑡𝑒𝑟 −5, 5 , 𝑅𝑖𝑔ℎ𝑡 10, 5 , 𝐿𝑒𝑓𝑡 −10, −5 , 𝑇𝑜𝑝 0, 8 , 𝐵𝑜𝑡𝑡𝑜𝑚 (0, −8) D. 𝐶𝑒𝑛𝑡𝑒𝑟 0, 00 , 𝑅𝑖𝑔ℎ𝑡 7, 3 , 𝐿𝑒𝑓𝑡 0, 5 , 𝑇𝑜𝑝 0, 8 , 𝐵𝑜𝑡𝑡𝑜𝑚 0, 1

SOLUTION 5: 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: 𝑥−ℎ 𝑎"

"

𝑦−𝑘 + 𝑏"

"

=1

The function of the ellipse we are evaluating is given as: 𝑥" 𝑦−3 + 49 4

"

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For the most part, the original equation is given to us in the standard form. It is important to note there is no value being subtracted from “x”, thus indicating ℎ = 0. 𝑥−0 49

"

𝑦−3 + 4

"

=1

We can then rewrite the function to reflect the squared terms in the denominator for 𝑎 𝑎𝑛𝑑 𝑏: (𝑥 − 0)" (𝑦 − 3)" + =1 7" 2" We now know that the values are 𝑎 = 7 and 𝑏 = 2 The point (ℎ, 𝑘) is the center of the ellipse, which is (0,3). The four extrema of an ellipse are calculated as: • Right Most Point: ℎ + 𝑎, 𝑘 = (7,3) • Left Most Point: ℎ − 𝑎, 𝑘 = (−7,3) • Top Most Point: ℎ, 𝑘 + 𝑏 = (0,5) • Bottom Most Point: ℎ, 𝑘 − 𝑏 = (0,1)

Therefore, the correct answer choice is D. 𝑪𝒆𝒏𝒕𝒆𝒓 𝟎, 𝟎 , 𝑹𝒊𝒈𝒉𝒕 𝟕, 𝟑 , 𝑳𝒆𝒇𝒕 −𝟕, 𝟑 , 𝑻𝒐𝒑 𝟎, 𝟓 , 𝑩𝒐𝒕𝒕𝒐𝒎 (𝟎, 𝟏)

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PROBLEM 6: Find the center and radius of the circle given by the equation: 𝑥 " − 6𝑥𝑦 + 𝑦 " + 10𝑦 + 27 = 0 A. 𝑟 = 2; ℎ, 𝑘 = (1, −2) B. 𝑟 = 3; ℎ, 𝑘 = (3, −5) C. 𝑟 = 4; ℎ, 𝑘 = (1, −15) D. 𝑟 = 5; ℎ, 𝑘 = (2, −9)

SOLUTION 6: 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.

Given the equation 𝑥 " − 6𝑥 + 𝑦 " + 10𝑦 + 27 = 0, the first step is to rearrange it so that it is in the standard form of a circle. The STANDARD FORM OF A CIRCLE centered at coordinates (ℎ, 𝑘) with radius “𝑟” is represented by the equation:

𝑟=

𝑥−ℎ

"

+ 𝑦−𝑘

"

or

𝑥−ℎ

"

+ 𝑦−𝑘

"

= 𝑟"

𝑥 " − 6𝑥 + 𝑦 " + 10𝑦 = −27



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We can then factor this equation: 𝑥 " − 6𝑥 + 9 + 𝑦 " + 10𝑦 + 27 = −27 + 9 + 27 After factoring and simplifying, we get: 𝑥−3

"

+ 𝑦+5

"

=9

With this equation now in the standard form, both the radius “𝑟”, and the center (ℎ. 𝑘) can be defined as: Radius 𝑟 = 3 Center (ℎ. 𝑘) = (3, −5)

Therefore, the correct answer choice is B. 𝒓 = 𝟑; 𝒉, 𝒌 = (𝟑, −𝟓)

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



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SOLUTION 7: 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. Solve the equations simultaneously by substituting the expression 𝑦 = 𝑥 − 1 into the equation of the circle such that: 𝑥" + 𝑥 − 1

"

−𝑥−3 𝑥−1 =0

𝑥 " + 𝑥 " − 2𝑥 + 1 − 𝑥 − 3𝑥 + 3 = 0 2𝑥 " − 6𝑥 + 4 = 0 𝑥 " − 3𝑥 + 2 = 0 Finally factoring: 𝑥−1 𝑥−2 =0 Which shows that the solutions where the equations will intersect and equal is 𝑥 = 1 or 𝑥 = 2. Plugging these values into the equation 𝑦 = 𝑥 − 1 to find the y=value of the intersection points we get: 𝑦 = 0 and 𝑦 = 1 Therefore, the points of intersection between the circle defined by 𝑥 " + 𝑦 " − 𝑥 − 3𝑦 = 0 and the line 𝑦 = 𝑥 − 1 are at (1,0) and 2,1 .

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

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PROBLEM 8: Find the center and radius of the circle given by the equation: 3𝑥 " + 3𝑦 " − 12𝑥 + 4 = 0

A. 𝑟 = B. 𝑟 =

{ |

; ℎ, 𝑘 = (2,0)

| "

; ℎ, 𝑘 = (3, −5)

C. 𝑟 = 4 2; ℎ, 𝑘 = (1, −15) D. 𝑟 = 5; ℎ, 𝑘 = (2, −9)

SOLUTION 8: 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.

Given the equation 3𝑥 " + 3𝑦 " − 12𝑥 + 4 = 0, the first step is to rearrange it so that it is in the standard form of a circle. The STANDARD FORM OF A CIRCLE centered at coordinates (ℎ, 𝑘) with radius “𝑟” is represented by the equation:

𝑟=



𝑥−ℎ

"

+ 𝑦−𝑘

"

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or

𝑥−ℎ

"

+ 𝑦−𝑘

"

= 𝑟"

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Collect the x and y terms, then divided through by 3 such that: }

𝑥 " − 4𝑥 + 𝑦 " + = 0 |

Subtract the constant from each side and complete the square on the x-term: }

𝑥 " − 4𝑥 + 4 + 𝑦 " = − + 4 |

Factoring and simplifying we get:

𝑥−2

"

+ 𝑦+0

"

=

{ |

With this equation now in the standard form, both the radius “𝑟”, and the center (ℎ. 𝑘) can be defined as:

Radius 𝑟 =

{ |



Center (ℎ. 𝑘) = (2,0)

Therefore, the correct answer choice is B. 𝒓 =



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𝟖 𝟑

; 𝒉, 𝒌 = (𝟐, 𝟎)

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

SOLUTION 9: 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: 𝐴𝑥 " + 𝐵𝑥𝑦 + 𝐶𝑦 " + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 In the general quadratic equation 8𝑦 " + 6𝑥𝑦 − 9 = 0, we find that: 𝐴 = 8, 𝐵 = 6, 𝑎𝑛𝑑 𝐶 = 0 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: 𝐵 " − 4𝐴𝐶 = 6

"

− 4 8 0 = 36

We know that when the 𝑑𝑖𝑠𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡 > 0, the conic section is a ℎ𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑎.

Therefore, the correct answer choice is B. 𝑯𝒚𝒑𝒆𝒓𝒃𝒐𝒍𝒂

PROBLEM 10: What is the radius of the circle defined by 𝑥 " + 𝑦 " − 4𝑥 + 8𝑦 = 7? A.

3

B. 2 5 C. 3 3 D. 4 3

SOLUTION 10: 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 general equation for a circle is given as: 𝑥−ℎ

"

+ 𝑦−𝑘

"

= 𝑟"

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As we are provided an equation that is not in this format, we can re-arrange it to make the standard form of the general equation. In order to do so, you must also complete the square: 𝑥 " − 4𝑥 + 4 + 𝑦 " + 8𝑦 + 16 = 7 + 4 + 16 We can then group the terms such they match the general equation: 𝑥−2

"

+ 𝑦+4

"

= 27

We can then solve for the radius of the circle: 𝑟 " = 27 𝑟 = 3√3

Therefore, the correct answer choice is C. 𝟑 𝟑.



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