We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

GRE Subject Test: Math : Hyperbolas

Study concepts, example questions & explanations for GRE Subject Test: Math

Create An Account

All GRE Subject Test: Math Resources

2 Diagnostic Tests 148 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #10 : Conic Sections

Using the information below, determine the equation of the hyperbola.

Foci: (-5,4) and (7,4)

Eccentricity: $\frac{3}{2}$

Possible Answers:

$\small \small \small \small \small \frac{(x-4)^{2}}{20} - \frac{(y-1)^{2}}{36}=1$

$\small \small \small \frac{(x-4)^{2}}{16} - \frac{(y-1)^{2}}{20}=1$

$\small \small \small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{36}=1$

$\small \small \small \frac{(x-1)^{2}}{20} - \frac{(y-4)^{2}}{16}=1$

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

Correct answer:

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

Explanation:

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity = c/a

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 12, we can set that equal to .

12=2c

$\small c= 6$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small c/a=3/2$

$\small \frac{c}{a}=\frac{3}{2}=\frac{6}{a}$

$\small 12=3a$

$\small a=4$

$\small \small a^{2}=16$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small 4^{2}+b^{2}=6^{2}$

$\small \small b^{2}=20$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 4, the center point will be on the same line. Hence, $\small k = 4$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 12, we can conclude that $\small h=1$

Center point: $\small \small (h, k)= (1,4)$

Thus, the equation of the hyperbola is:

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

Report an Error

Example Question #1 : Hyperbolas

Using the information below, determine the equation of the hyperbola.

Foci: (1, 8) and (9, 8)

Eccentricity:

Possible Answers:

$\small \small \small \small \small \small \small \frac{(x-5)^{2}}{12} - \frac{(y-8)^{2}}{4}=1$

$\small \small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{16}=1$

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

$\small \small \small \small \small \small \small \small \frac{(x-8)^{2}}{12} - \frac{(y-5)^{2}}{16}=1$

$\small \small \small \small \small \small \small \frac{(x-8)^{2}}{4} - \frac{(y-5)^{2}}{12}=1$

Correct answer:

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

Explanation:

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity = c/a

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 8, we can set that equal to .

$\small 8=2c$

$\small \small c= 4$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small \small \small c/a=2$

$\small \small \frac{c}{a}=\frac{2}{1}=\frac{4}{a}$

$\small \small 4=2a$

$\small \small a=2$

$\small \small \small a^{2}=4$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small \small 2^{2}+b^{2}=4^{2}$

$\small \small \small b^{2}=12$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 8, the center point will be on the same line. Hence, $\small \small k = 8$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 8, we can conclude that $\small \small h=5$

Center point: $\small \small \small (h, k)= (5, 8 )$

Thus, the equation of the hyperbola is:

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

Report an Error

Example Question #4 : Conic Sections

Find the coordinate of intersection, if possible: y=x^2+4 and y=-x^2+4 .

Possible Answers:

(4,0)

$(0,\pm4)$

(0,4)

$\textup{There is no intersection.}$

(-4,0)

Correct answer:

(0,4)

Explanation:

To solve for x and y, set both equations equal to each other and solve for x.

-x^2+4=x^2+4

2x^2 =0

x=0

Substitute x=0 into either parabola.

y=0^2+4 = 4

The coordinate of intersection is (0,4) .

Report an Error

Example Question #5 : Conic Sections

Find the intersection(s) of the two parabolas: y=2x^2+3x+4 , y=-4x^2+4

Possible Answers:

$\left(\frac{1}{2},3\right)$

$\left(-\frac{1}{2},3\right)$

$(0,4) \textup{ and }\left(-\frac{1}{2},3\right)$

(0,4)

$(0,4) \textup{ and }\left(\frac{1}{2},3\right)$

Correct answer:

$(0,4) \textup{ and }\left(-\frac{1}{2},3\right)$

Explanation:

Set both parabolas equal to each other and solve for x.

2x^2+3x+4=-4x^2+4

6x^2+3x=0

3x(2x+1)=0

$x=0,-\frac{1}{2}$

Substitute both values of into either parabola and determine .

y=-4(0)^2+4= 4

$y=-4\left(-\frac{1}{2}\right)^2+4= -4\left(\frac{1}{4}\right)+4= -1+4=3$

The coordinates of intersection are:

(0,4) and $\left(-\frac{1}{2},3\right)$

Report an Error

Example Question #6 : Conic Sections

Find the points of intersection:

y = x^2 + 40x - 6 ; y = -x^2 + 19 x + 5

Possible Answers:

(11, 555) ; (0.5 , 14.25)

(-11, -325); (0.5, 14.25)

(-11, -567); (0.5, 9.05)

(-0.5, -25.75) ; (-11, -325)

(-11, -567) ; (-0.5, -25.75)

Correct answer:

(-11, -325); (0.5, 14.25)

Explanation:

To solve, set both equations equal to each other:

x^2 + 40x - 6 = -x^2 + 19x + 5

To solve as a quadratic, combine like terms by adding/subtracting all three terms from the right side to the left side:

x^2 + x^2 +40x - 19x - 6 - 5 = 0

This simplifies to

2x^2 + 21 x - 11 = 0

Solving by factoring or the quadratic formula gives the solutions and 0.5 .

Plugging each into either original equation gives us:

(-11)^2 + 40(-11 ) - 6 = 121 - 440 - 6 = -325

(0.5) ^ 2 + 40 (0.5) - 6 = 14.25

Our coordinate pairs are (-11, -325) and (0.5, 14.25 ) .

Report an Error

View Tutors

Mohammed
Certified Tutor

Undergradute-University of Maryland-College Park, Bachelor of Science, Electrical Engineering. Graduate Degree-University of ...

View Tutors

Jay
Certified Tutor

University of Windsor, Master's/Graduate, Management with a specialisation in Business Data Analytics.

View Tutors

Aja
Certified Tutor

Western Governors University, Bachelor in Arts, Mathematics Teacher Education.

All GRE Subject Test: Math Resources

2 Diagnostic Tests 148 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Algebra Tutors in Philadelphia, LSAT Tutors in Los Angeles, English Tutors in New York City, Reading Tutors in Washington DC, SAT Tutors in San Francisco-Bay Area, English Tutors in Seattle, Calculus Tutors in Phoenix, Calculus Tutors in San Francisco-Bay Area, ISEE Tutors in Atlanta, Chemistry Tutors in Seattle

Popular Courses & Classes

MCAT Courses & Classes in Atlanta, GMAT Courses & Classes in San Diego, ACT Courses & Classes in Washington DC, GMAT Courses & Classes in Houston, GRE Courses & Classes in Los Angeles, ACT Courses & Classes in New York City, ACT Courses & Classes in Denver, ISEE Courses & Classes in Denver, SSAT Courses & Classes in Boston, LSAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

ACT Test Prep in New York City, GMAT Test Prep in San Diego, ACT Test Prep in Seattle, MCAT Test Prep in Dallas Fort Worth, SAT Test Prep in New York City, SAT Test Prep in Miami, SSAT Test Prep in Philadelphia, LSAT Test Prep in Dallas Fort Worth, ISEE Test Prep in Boston, GRE Test Prep in Washington DC

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

GRE Subject Test: Math : Hyperbolas

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #10 : Conic Sections

Example Question #1 : Hyperbolas

Example Question #4 : Conic Sections

Example Question #5 : Conic Sections

Example Question #6 : Conic Sections

All GRE Subject Test: Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: