Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Conic Sections

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 #62 : Understand Features Of Hyperbolas And Ellipses

Find the equations of the asymptotes of the hyperbola with the following equation:

$\frac{x^2}{36}-\frac{y^2}{49}=1$

Possible Answers:

$y=\pm\frac{6}{7}x$

$y=\pm\frac{49}{36}x$

$y=\pm\frac{7}{6}$

$y=\pm\frac{7}{6}x$

Correct answer:

$y=\pm\frac{7}{6}x$

Explanation:

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , where (h,k) is the center of the hyperbola.

The equations of the asymptotes for this hyperbola are given by the following equations:

$y=\pm\frac{b}{a}x$

For the hyperbola in question, a=6 and b=7 .

Thus, the equations for its asymptotes are $y=\pm\frac{7}{6}x$

Report an Error

Example Question #162 : Conic Sections

Find the equations of the asymptotes for the hyperbola with the following equation:

4x^2-25y^2+24x+550y-3089=0

Possible Answers:

$y=\pm\frac{25}{4}x$

$y=\frac{5}{2}x-\frac{5}{2}x$

$y=-\frac{5}{2}x+\frac{9}{2}$

$y=\frac{2}{5}x+\frac{61}{5}$

$y=-\frac{2}{5}x+\frac{49}{5}$

$y=\frac{3}{11}x+\frac{67}{11}$

$y=-\frac{3}{11}x+\frac{50}{11}$

Correct answer:

$y=\frac{2}{5}x+\frac{61}{5}$

$y=-\frac{2}{5}x+\frac{49}{5}$

Explanation:

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , where (h,k) is the center of the hyperbola.

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

Group the terms together and terms together.

4x^2-25y^2+24x+550y-3089=0

4x^2+24x-25y^2+550y-3089=0

Factor out from the terms and from the terms.

4(x^2+6x)-25(y^2-22y)-3089=0

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

4(x^2+6x+9)-25(y^2-22y+121)-3089=-2989

Add 3089 to both sides of the equation:

4(x^2+6x+9)-25(y^2-22y+121)=100

Divide both sides by 100 .

$\frac{x^2+6x+9}{25}-\frac{y^2-22y+121}{4}=1$

Factor the two terms to get the standard form of the equation of a hyperbola.

$\frac{(x+3)^2}{25}-\frac{(y-11)^2}{4}=1$

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, a=5 and b=2 .

Thus, the slopes for its asymptotes are $\pm\frac{2}{5}$ .

Now, use the center of the hyperbola, (-3, 11) , to plug into the point-slope form of a line to find the equations of the asymptotes.

The first asymptote has the following equation:

$y-11=\frac{2}{5}(x+3)$

$y-11=\frac{2}{5}x+\frac{6}{5}$

$y=\frac{2}{5}x+\frac{61}{5}$

The second asymptote has the following equation:

$y-11=-\frac{2}{5}(x+3)$

$y-11=-\frac{2}{5}x-\frac{6}{5}$

$y=-\frac{2}{5}x+\frac{49}{5}$

Report an Error

Example Question #163 : Conic Sections

Find the equations of the asymptotes for the hyperbola with the following equation:

9x^2-4y^2-162x-8y+689=0

Possible Answers:

$y=\frac{3}{2}x-\frac{29}{2}$

$y=-\frac{3}{2}x-\frac{25}{2}$

$y=\frac{9}{4}x-\frac{14}{4}$

$y=-\frac{9}{4}x+\frac{3}{4}$

$y=\frac{1}{9}x-2$

$y=-\frac{1}{9}x+\frac{16}{9}$

$y=\frac{2}{3}x+\frac{4}{3}$

$y=-\frac{2}{3}x-2$

Correct answer:

$y=\frac{3}{2}x-\frac{29}{2}$

$y=-\frac{3}{2}x-\frac{25}{2}$

Explanation:

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , where (h,k) is the center of the hyperbola.

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

Group the terms together and terms together.

9x^2-4y^2-162x-8y+689=0

9x^2-162x-4y^2-8y+689=0

Factor out from the terms and from the terms.

9(x^2-18x)-4(y^2+2y)+689=0

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

9(x^2-18x+81)-4(y^2+2y+1)+689=725

Subtract 689 from both sides of the equation:

9(x^2-18x+81)-4(y^2+2y+1)=36

Divide both sides by .

$\frac{x^2-18x+81}{4}-\frac{y^2+2y+1}{9}=1$

Factor the two terms to get the standard form of the equation of a hyperbola.

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

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, a=2 and b=3 .

Thus, the slopes for its asymptotes are $m=\pm\frac{3}{2}$ .

Plug in the center of the hyperbola into the point-slope form of a line to find the equations of the asymptotes. The center of the hyperbola is (9, -1) .

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

Now, simplify each equation. For the first equation,

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

$y+1=\frac{3}{2}x-\frac{27}{2}$

$y=\frac{3}{2}x-\frac{29}{2}$

For the second equation,

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

$y+1=-\frac{3}{2}x+\frac{27}{2}$

$y=-\frac{3}{2}x-\frac{25}{2}$

Report an Error

Example Question #1741 : Pre Calculus

Find the equations of the asymptotes for the hyperbola with the following equation:

$\frac{(x-12)^2}{144}-\frac{(y+\sqrt3)^2}{169}=1$

Possible Answers:

$y=\pm\frac{169}{144}(x+12)-\sqrt3$

$y=\pm\frac{144}{169}(x-12)-\sqrt3$

$y=\pm\frac{12}{13}(x+\sqrt3)+12$

$y=\pm\frac{13}{12}(x-12)+\sqrt3$

Correct answer:

$y=\pm\frac{13}{12}(x-12)+\sqrt3$

Explanation:

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , where (h,k) is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, a=12 and b=13 .

Thus, the slopes for its asymptotes are $m=\pm\frac{13}{12}$

Now, plug in the coordinates for the center of the hyperbola $(12, \sqrt3)$ into the point-slope form of the line to find the equations of the asymptotes.

$y-\sqrt3=\pm\frac{13}{12}(x-12)$

$y=\pm\frac{13}{12}(x-12)+\sqrt3$

Report an Error

Example Question #1742 : Pre Calculus

Find the equations of the asymptotes for the hyperbola with the following equation:

$\frac{(x-2)^2}{100}-\frac{(y-6)^2}{36}=1$

Possible Answers:

$y=\frac{3}{5}x+\frac{24}{5}$

$y=-\frac{3}{5}x+\frac{36}{5}$

$y=\frac{5}{3}x-\frac{7}{3}$

$y=-\frac{5}{3}x+\frac{8}{3}$

$y=\frac{25}{9}x-\frac{8}{9}$

$y=-\frac{25}{9}x+\frac{2}{3}$

$y=\frac{9}{25}x-\frac{1}{5}$

$y=-\frac{9}{25}x+\frac{9}{5}$

Correct answer:

$y=\frac{3}{5}x+\frac{24}{5}$

$y=-\frac{3}{5}x+\frac{36}{5}$

Explanation:

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , where (h,k) is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, and b=6 .

Thus, the slopes for its asymptotes are $m=\pm\frac{6}{10}=\pm\frac{3}{5}$ .

Now, use the point-slope form of a line in addition to the center of the hyperbola to find the equations of the asymptotes.

The center is at (2, 6) .

The first equation of the asymptote would be the following:

$y-6=\frac{3}{5}(x-2)$

$y-6=\frac{3}{5}x-\frac{6}{5}$

$y=\frac{3}{5}x+\frac{24}{5}$

The second equation of the asymptote would be the following:

$y-6=-\frac{3}{5}(x-2)$

$y-6=-\frac{3}{5}x+\frac{6}{5}$

$y=-\frac{3}{5}+\frac{36}{5}$

Report an Error

Example Question #1743 : Pre Calculus

Find the equations of the asymptotes for the hyperbola with the following equation:

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

Possible Answers:

$y=\pm\frac{9}{25}x$

$y=\pm\frac{3}{5}x$

$y=\pm\frac{5}{3}x$

$y=\pm\frac{25}{9}x$

Correct answer:

$y=\pm\frac{3}{5}x$

Explanation:

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where (h,k) is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, a=5 and b=3 .

Since the center is (0, 0) , the equations for its asymptotes are $y=\pm\frac{3}{5}x$ .

Report an Error

Example Question #1744 : Pre Calculus

Find the equations of the asymptotes for the hyperbola with the following equation:

$\frac{(y-9)^2}{169}-\frac{(x-2)^2}{225}=1$

Possible Answers:

y=2x-9

y=-2x+8

$y=\frac{15}{13}x-\frac{20}{13}$

$y=-\frac{15}{13}x+\frac{5}{13}$

$y=\frac{13}{15}x+\frac{109}{15}$

$y=-\frac{13}{15}x+\frac{161}{15}$

$y=\frac{2}{9}x-\frac{7}{2}$

$y=-\frac{2}{9}x+\frac{25}{2}$

Correct answer:

$y=\frac{13}{15}x+\frac{109}{15}$

$y=-\frac{13}{15}x+\frac{161}{15}$

Explanation:

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where (h,k) is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, a=15 and b=13 .

Thus, the slopes for its asymptotes are $\pm\frac{13}{15}$ .

Now, plug in the center of the hyperbola, (2, 9) into the point-slope form of a line to find the equations of the asymptotes.

For the first asymptote,

$y-9=\frac{13}{15}(x-2)$

$y-9=\frac{13}{15}x-\frac{26}{15}$

$y=\frac{13}{15}x+\frac{109}{15}$

For the second asymptote,

$y-9=-\frac{13}{15}(x-2)$

$y-9=-\frac{13}{15}x+\frac{26}{15}$

$y=-\frac{13}{15}x+\frac{161}{15}$

Report an Error

Example Question #1745 : Pre Calculus

Find the equations of the asymptotes for the hyperbola with the following equation:

$\frac{(y-1)^2}{8}-\frac{(x+2)^2}{36}=1$

Possible Answers:

$y=\pm\frac{3\sqrt2}{2}(x-2)-1$

$y=\pm\frac{\sqrt2}{3}(x+2)+1$

$y=\pm\frac{1}{8}(x-2)-1$

$y=\pm\frac{2\sqrt2}{3}(x-1)+9$

Correct answer:

$y=\pm\frac{\sqrt2}{3}(x+2)+1$

Explanation:

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where (h,k) is the center of the hyperbola.

The slopes of this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, a=6 and $b=\sqrt8=2\sqrt2$ .

Thus, the slopes for its asymptotes are $\pm\frac{2\sqrt2}{6}=\pm\frac{\sqrt2}{3}$ .

Now, plug in the center of the hyperbola into the point-slope form of the equation fo the line to get the equations of the asymptotes.

The center of the hyperbola is (-2, 1) . The equations of the asymptotes are then:

$y-1=\pm\frac{\sqrt2}{3}(x+2)$

$y=\pm\frac{\sqrt2}{3}(x+2)+1$

Report an Error

Example Question #1751 : Pre Calculus

Find the equations of the asymptotes for the hyperbola with the following equation:

$\frac{(y-4)^2}{196}-\frac{(x+\pi)^2}{400}=1$

Possible Answers:

$y=\pm\frac{13}{20}(x-\pi)+8$

$y=\pm\frac{13}{20}(x+\pi)+4$

$y=\pm\frac{\pi}{4}(x-2)+1$

$y=\pm\frac{20}{13}(x+4)+\pi$

Correct answer:

$y=\pm\frac{13}{20}(x+\pi)+4$

Explanation:

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where (h,k) is the center of the hyperbola.

The slopes of this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, a=20 and b=13 .

Thus, the slopes for its asymptotes are $\pm\frac{13}{20}$ .

Now, plug in the center of the hyperbola into the point-slope form of an equation of a line to find the equations of the asymptotes. The center of the hyperbola is $(-\pi, 4)$ .

The equations of the asymptotes can be then given by the following:

$y-4=\pm\frac{13}{20}(x+\pi)$

$y=\pm\frac{13}{20}(x+\pi)+4$

Report an Error

Example Question #6 : Hyperbolas

Find the equations of the asymptotes for the hyperbola with the following equation:

36y^2-16x^2-144y-32x-448=0

Possible Answers:

$y=\frac{3}{2}x+3$

$y=-\frac{3}{2}x-\frac{5}{2}$

$y=\pm\frac{4}{3}x$

$y=\frac{1}{2}x+3$

$y=-\frac{1}{2}x+1$

$y=\frac{2}{3}x+\frac{8}{3}$

$y=-\frac{2}{3}x+\frac{4}{3}$

Correct answer:

$y=\frac{2}{3}x+\frac{8}{3}$

$y=-\frac{2}{3}x+\frac{4}{3}$

Explanation:

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where (h,k) is the center of the hyperbola.

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

Group the terms together and terms together.

36y^2-16x^2-144y-32x-448=0

36y^2-144y-16x^2-32x-448=0

Factor out from the terms and from the terms.

36(y^2-4y)-16(x^2+2x)-448=0

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

36(y^2-4y+4)-16(x^2+2x+1)-448=98

Add 448 to both sides of the equation:

36(y^2-4y+4)-16(x^2+2x+1)=576

Divide both sides by 576 .

$\frac{y^2-4y+4}{16}-\frac{x^2+2x+1}{36}=1$

Factor the two terms to get the standard form of the equation of a hyperbola.

$\frac{(y-2)^2}{16}-\frac{(x+1)^2}{36}=1$

The slopes of this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, a=6 and b=4 .

Thus, the slopes for its asymptotes are $\pm\frac{4}{6}=\pm\frac{2}{3}$ .

Now, plug in the center of the hyperbola, (-1, 2) into the point-slope form of the equation of a line to get the equations of the asymptotes.

For the first equation,

$y-2=\frac{2}{3}(x+1)$

$y-2=\frac{2}{3}x+\frac{2}{3}$

$y=\frac{2}{3}x+\frac{8}{3}$

For the second equation,

$y-2=-\frac{2}{3}(x+1)$

$y-2=-\frac{2}{3}x-\frac{2}{3}$

$y=-\frac{2}{3}x+\frac{4}{3}$

Report an Error

View Pre-Calculus Tutors

Siddharth
Certified Tutor

University of Petroleum and Energy Studies and, Masters in Business Administration, Aviation Management.

View Pre-Calculus Tutors

Tracy
Certified Tutor

Clarion University of Pennsylvania, Bachelor of Science, Physics.

View Pre-Calculus Tutors

Cassandra
Certified Tutor

Marygrove College, Master of Arts, High School Teaching.

All Precalculus Resources

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

Popular Subjects

Computer Science Tutors in Atlanta, ACT Tutors in Dallas Fort Worth, Algebra Tutors in Denver, GRE Tutors in Los Angeles, ISEE Tutors in Houston, Algebra Tutors in Houston, Calculus Tutors in San Francisco-Bay Area, SSAT Tutors in Philadelphia, English Tutors in New York City, MCAT Tutors in San Francisco-Bay Area

Popular Courses & Classes

ACT Courses & Classes in San Diego, MCAT Courses & Classes in San Diego, SSAT Courses & Classes in New York City, SSAT Courses & Classes in Washington DC, LSAT Courses & Classes in San Diego, ACT Courses & Classes in Chicago, LSAT Courses & Classes in Denver, SAT Courses & Classes in New York City, ISEE Courses & Classes in Houston, GRE Courses & Classes in Denver

Popular Test Prep

GMAT Test Prep in Chicago, LSAT Test Prep in Washington DC, LSAT Test Prep in Boston, SAT Test Prep in Miami, LSAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Chicago, GMAT Test Prep in New York City, GMAT Test Prep in Los Angeles, ACT Test Prep in San Diego, SSAT Test Prep in New York City

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 : Conic Sections

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #62 : Understand Features Of Hyperbolas And Ellipses

Example Question #162 : Conic Sections

Example Question #163 : Conic Sections

Example Question #1741 : Pre Calculus

Example Question #1742 : Pre Calculus

Example Question #1743 : Pre Calculus

Example Question #1744 : Pre Calculus

Example Question #1745 : Pre Calculus

Example Question #1751 : Pre Calculus

Example Question #6 : Hyperbolas

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: