We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Understand features of hyperbolas and ellipses

Study concepts, example questions & explanations for Precalculus

Create An Account

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #181 : Conic Sections

Find the vertices of the hyperbola with the following equation:

4y^2-25x^2-56y-150x-129=0

Possible Answers:

(-1, 7)(-5, 7)

(9, 1)(-1, 9)

(-3, 12)(-3, 2)

(12, 3)(12, -3)

Correct answer:

(-3, 12)(-3, 2)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the vertices are located at (h-a, k)(h+a, k) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the vertices are located at (h, k+b)(h, k-b) .

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the terms together and the terms together.

4y^2-25x^2-56y-150x-129=0

4y^2-56y-25x^2-150x-129=0

Factor out from the terms and from the terms.

4(y^2-14y)-25(x^2+6x)-129=0

Now complete the squares. Remember to add the same amount to both sides of the equation!

4(y^2-14y+49)-25(x^2+6x+9)-129=-29

Add 129 to both sides of the equation.

4(y^2-14y+49)-25(x^2+6x+9)=100

Factor the square portions of the equation.

4(y-7)^2-25(x+3)^2=100

Divide both sides by 100 to get the standard form for the equation of a hyperbola.

$\frac{(y-7)^2}{25}-\frac{(x+3)^2}{4}=1$

For the hyperbola in question, the center is located at (-3, 7) and b=5 . The vertices must be at (-3, 12) and (-3, 2) .

Report an Error

Example Question #92 : Understand Features Of Hyperbolas And Ellipses

Find the vertices of the hyperbola with the following equation:

100y^2-225x^2-2000y+450x-12725=0

Possible Answers:

(11, 10)(-9, 10)

(1, 25)(1, -5)

(16, 10)(-14, 10)

(1, 20)(1, 0)

Correct answer:

(1, 25)(1, -5)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the vertices are located at (h-a, k)(h+a, k) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the vertices are located at (h, k+b)(h, k-b) .

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the terms together and the terms together.

100y^2-225x^2-2000y+450x-12725=0

100y^2-2000y-225x^2+450x-12725=0

Factor out 100 from the terms and from the terms.

100(y^2-20y)-225(x^2-2x)-12725=0

From here we need to complete the squares. Remember to add the same amount to both sides of the equation!

100(y^2-20y+100)-225(x^2-2x+1)-12725=9775

Add 12725 to both sides of the equation.

100(y^2-20y+100)-225(x^2-2x+1)=22500

Factor the square portions of the equation.

100(y-10)^2-225(x-1)^2=22500

Divide both sides by 22500 to get the standard form for the equation of a hyperbola.

$\frac{(y-10)^2}{225}-\frac{(x-1)^2}{100}=1$

For the hyperbola in question, the center is located at (1, 10) and b=15 . The vertices must be at (1, 25) and (1, -5) .

Report an Error

Example Question #93 : Understand Features Of Hyperbolas And Ellipses

Find the endpoints of the conjugate axis of the hyperbola with the following equation:

4x^2-16y^2-16x+64y-112=0

Possible Answers:

(2, 4)(2, 0)

(0, 2)(4, 2)

(4, 4)(0, 0)

(2, 4)(2, -4)

Correct answer:

(2, 4)(2, 0)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the endpoints of the conjugate axis are located at (h, k+b)(h, k-b) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the endpoints of the conjugate axis are located at (h-a, k)(h+a, k) .

Start by putting the given equation into the standard form for the equation of a hyperbola.

Group the terms together and the terms together.

4x^2-16y^2-16x+64y-112=0

4x^2-16x-16y^2+64y-112=0

Factor out from the terms and from the terms.

4(x^2-4x)-16(y^2-4y)-112=0

Complete the squares. Remember to add the same amount to both sides of the equation!

4(x^2-4x+4)-16(y^2-4y+4)-112=-48

Add 112 to both sides of the equation.

4(x^2-4x+4)-16(y^2-4y+4)=64

Factor the squares.

4(x-2)^2-16(y-2)^2=64

Divide both sides by to get the standard form for the equation of a hyperbola.

$\frac{(x-2)^2}{16}-\frac{(y-2)^2}{4}=1$

For the hyperbola in question, the center is located at (2,2) and b=2 . The endpoints of the conjugate axis must be at (2, 0) and (2, 4) .

Report an Error

Example Question #94 : Understand Features Of Hyperbolas And Ellipses

Find the endpoints of the conjugate axis of the hyperbola with the following equation:

9y^2-4x^2-216y+88x+776=0

Possible Answers:

(12, 14)(12, 8)

(11, 13)(11, 9)

(12, 9)(12, 13)

(14,12)(8, 12)

Correct answer:

(14,12)(8, 12)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ and $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the endpoints of the conjugate axis are located at (h, k+b)(h, k-b) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the endpoints of the conjugate axis are located at (h-a, k)(h+a, k) .

Start by putting the given equation into the standard form for the equation of a hyperbola.

Group the terms together and the terms together.

9y^2-4x^2-216y+88x+776=0

9y^2-216y-4x^2+88x+776=0

Factor out from the terms and from the terms.

9(y^2-24y)-4(x^2-22x)+776=0

Complete the squares. Remember to add the same amount to both sides of the equation!

9(y^2-24y+144)-4(x^2-22x+121)+776=812

Subtract 776 from both sides of the equation.

9(y^2-24y+144)-4(x^2-22x+121)=36

Factor the squares.

9(y-12)^2-4(x-11)^2=36

Divide both sides by to get the standard form for the equation of a hyperbola.

$\frac{(y-12)^2}{4}-\frac{(x-11)^2}{9}=1$

For the hyperbola in question, the center is located at (11, 12) and a=3 . The endpoints of the conjugate axis must be at (14, 12) and (8, 12) .

Report an Error

Example Question #91 : Hyperbolas And Ellipses

Find the eccentricity of the following hyperbola: $\frac{x^2}{4}-\frac{y^2}{25}=1$

Possible Answers:

$\frac{\sqrt{41}}{5}$

$\frac{\sqrt{41}}{2}$

$\frac{\sqrt{29}}{5}$

$\frac{\sqrt{29}}{2}$

$\sqrt{34}$

Correct answer:

$\frac{\sqrt{29}}{2}$

Explanation:

In order to find the eccentricity of $\frac{x^2}{4}-\frac{y^2}{25}=1$ , first determine the values of and from the standard form of the hyperbola:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

a^2=4

a=2

b^2=25

b=5

Use the following formula to calculate eccentricity. The eccentricity of a hyperbola should always be greater than 1.

$e=\frac{\sqrt{a^2+b^2}}{a}$

Substitute and solve for eccentricity.

$e=\frac{\sqrt{2^2+5^2}}{2}=\frac{\sqrt{29}}{2}$

Report an Error

Example Question #92 : Hyperbolas And Ellipses

Find the eccentricity of the hyperbola with the following equation:

$\frac{x^2}{144}-\frac{y^2}{25}=1$

Possible Answers:

$\frac{12}{13}$

$\frac{25}{144}$

$\frac{144}{25}$

$\frac{13}{12}$

Correct answer:

$\frac{13}{12}$

Explanation:

Recall the standard form of the equation of a horizontal hyperbola:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$ , where (h,k) is the center.

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{a}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

a=12 and b=5

$c=\sqrt{144+25}=\sqrt{169}=13$

Thus,

$e=\frac{13}{12}$

Report an Error

Example Question #93 : Hyperbolas And Ellipses

Find the eccentricity of the following hyperbola:

$\frac{(x+2)^2}{9}-\frac{(y-1)^2}{25}=1$

Possible Answers:

$\frac{3\sqrt{34}}{3}$

$\frac{5}{3}$

$\frac{9}{25}$

$\frac{\sqrt{34}}{3}$

Correct answer:

$\frac{\sqrt{34}}{3}$

Explanation:

Recall the standard form of the equation of a horizontal hyperbola:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$ , where (h,k) is the center

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{a}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

a=3 and b=5

$c=\sqrt{9+25}=\sqrt{34}$

Thus,

$e=\frac{\sqrt{34}}{3}$

Report an Error

Example Question #94 : Hyperbolas And Ellipses

Find the eccentricity of the following hyperbola:

$\frac{(x+2)^2}{13}-\frac{(y-9)^2}{12}=1$

Possible Answers:

$\frac{5\sqrt{13}}{13}$

$\frac{5\sqrt{13}}{12}$

$\frac{5}{13}$

$\frac{13}{12}$

Correct answer:

$\frac{5\sqrt{13}}{13}$

Explanation:

Recall the standard form of the equation of a horizontal hyperbola:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$ , where (h,k) is the center

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{a}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

$a=\sqrt{13}$ and $b=\sqrt{12}$

$c=\sqrt{13+12}=\sqrt{25}=5$

Thus,

$e=\frac{5}{\sqrt{13}}=\frac{5\sqrt{13}}{13}$

Report an Error

Example Question #95 : Hyperbolas And Ellipses

Find the eccentricity of the following hyperbola:

$\frac{(x-3)^2}{21}-\frac{(y+9)^2}{28}=1$

Possible Answers:

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{\sqrt3}{21}$

$\frac{\sqrt{21}}{3}$

Correct answer:

$\frac{\sqrt{21}}{3}$

Explanation:

Recall the standard form of the equation of a horizontal hyperbola:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$ , where (h,k) is the center

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{a}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

$a=\sqrt{21}$ and $b=\sqrt{28}$

$c=\sqrt{21+28}=\sqrt{49}=7$

Thus,

$e=\frac{7}{\sqrt{21}}=\frac{7\sqrt{21}}{21}=\frac{\sqrt{21}}{3}$

Report an Error

Example Question #191 : Conic Sections

Find the eccentricity of the following hyperbola:

$\frac{(x-3)^1}{100}-\frac{(y+4)^2}{21}=1$

Possible Answers:

$\frac{11}{10}$

$\frac{10}{11}$

$\frac{100}{11}$

$\frac{121}{100}$

Correct answer:

$\frac{11}{10}$

Explanation:

Recall the standard form of the equation of a horizontal hyperbola:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$ , where (h,k) is the center

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{a}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

a=10 and $b=\sqrt{21}$

$c=\sqrt{100+21}=\sqrt{121}=11$

Thus,

$e=\frac{11}{10}$

Report an Error

View Pre-Calculus Tutors

Philipp
Certified Tutor

University of Nevada-Las Vegas, Bachelor of Science, Mechanical Engineering.

View Pre-Calculus Tutors

Gary
Certified Tutor

SUNY at Albany, Bachelor of Science, Economics. Pace University-New York, Masters in Business Administration, Analysis and Fu...

View Pre-Calculus Tutors

Erik Weijie
Certified Tutor

University of Florida, Bachelor of Science, Chemical Engineering.

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

LSAT Tutors in San Diego, Algebra Tutors in San Francisco-Bay Area, Calculus Tutors in Los Angeles, ISEE Tutors in Washington DC, Statistics Tutors in Dallas Fort Worth, Statistics Tutors in San Diego, French Tutors in Philadelphia, Computer Science Tutors in New York City, LSAT Tutors in Boston, GMAT Tutors in Washington DC

Popular Courses & Classes

SSAT Courses & Classes in Phoenix, SAT Courses & Classes in Seattle, ACT Courses & Classes in Chicago, GMAT Courses & Classes in Seattle, Spanish Courses & Classes in Houston, Spanish Courses & Classes in Miami, LSAT Courses & Classes in Phoenix, GRE Courses & Classes in New York City, GRE Courses & Classes in Phoenix, LSAT Courses & Classes in New York City

Popular Test Prep

LSAT Test Prep in San Diego, SAT Test Prep in Los Angeles, GRE Test Prep in Phoenix, SAT Test Prep in Denver, SAT Test Prep in Philadelphia, SAT Test Prep in Dallas Fort Worth, SAT Test Prep in Phoenix, GRE Test Prep in New York City, SAT Test Prep in Boston, GMAT Test Prep in Dallas Fort Worth

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Precalculus : Understand features of hyperbolas and ellipses

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #181 : Conic Sections

Example Question #92 : Understand Features Of Hyperbolas And Ellipses

Example Question #93 : Understand Features Of Hyperbolas And Ellipses

Example Question #94 : Understand Features Of Hyperbolas And Ellipses

Example Question #91 : Hyperbolas And Ellipses

Example Question #92 : Hyperbolas And Ellipses

Example Question #93 : Hyperbolas And Ellipses

Example Question #94 : Hyperbolas And Ellipses

Example Question #95 : Hyperbolas And Ellipses

Example Question #191 : Conic Sections

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: