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

← Previous 1 2 3 4 5 6 7 8 9 … 12 13 Next →

Example Question #1 : Understand Features Of Hyperbolas And Ellipses

Consider the following equation:

9x^2 - 18x + 100y^2 + 600y = -9

Classify the equation by its form when graphed.

Possible Answers:

A hyperbola with foci at (3, -1) and (1, 3)

A hyperbola with foci at (10, 3) and (-10, -3)

A circle centered at (3, -1)

An ellipse centered at (1, -3)

An ellipse centered at (10, 3)

Correct answer:

An ellipse centered at (1, -3)

Explanation:

Start with the equation

9x^2 - 18x + 100y^2 + 600y = -9

It would be helpful if we "complete the square" for both the x and y polynomials.

So by adding 9 and 900 we transform the equation into

And then by dividing 900, we see that the outcome is:

$\frac{(x-1)^2}{100} + \frac{(y+3)^2}{9} = 1$

The standard equation for an ellipse is

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

With a center at (h, k)

Therefore this is the equation of an ellipse, with center (1, -3)

Report an Error

Example Question #2 : Understand Features Of Hyperbolas And Ellipses

Express the following equation for an ellipse in standard form:

4(x-4)^2+9(x-7)^2=36

Possible Answers:

$\frac{4(x-4)^2}{36}+\frac{9(y-7)^2}{36}=36$

$\frac{(x-4)^2}{9}+\frac{(y-7)^2}{4}=36$

4(x-4)^2+9(y-7)^2=1

$\frac{(x-4)^2}{9}+\frac{(y-7)^2}{4}=1$

(x-4)^2+(y-7)^2=1

Correct answer:

$\frac{(x-4)^2}{9}+\frac{(y-7)^2}{4}=1$

Explanation:

Remember that the equation for an ellipse in standard form looks like the following:

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

Where the point (h,k) gives the center of the ellipse, a is half the length of its axis in the x direction, and b is half the length of its axis in the y direction. We can see that this form has a 1 on the right side of the equation, so let's start by dividing both sides of our equation by 36 to get a 1 on the right side:

4(x-4)^2+9(x-7)^2=36

$\frac{4(x-4)^2+9(x-7)^2}{36}=\frac{36}{36}$

$\frac{4(x-4)^2}{36}+\frac{9(x-7)^2}{36}=1$

Now we can simplify the fractions on the right side of the equation, which gives us the equation for our ellipse in standard form:

$\frac{(x-4)^2}{9}+\frac{(y-7)^2}{4}=1$

This ellipse would have its center at (4,7), would be 6 units wide in the x direction, and 4 units wide in the y direction, because a^2=9 so a=3 , and b^2=4 so b=2 .

Report an Error

Example Question #3 : Understand Features Of Hyperbolas And Ellipses

Which of the following is an equation for an ellipse written in standard form?

Possible Answers:

$\frac{(x+3)^2}{16}-\frac{(y+4)^2}{25}=1$

$\frac{(y-3)^2}{4}+\frac{(x+5)^2}{16}=1$

$\frac{(y-9)^2}{25}-\frac{(x-10)^2}{9}=1$

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

(y+2)^2+(x-4)^2=81

Correct answer:

$\frac{(y-3)^2}{4}+\frac{(x+5)^2}{16}=1$

Explanation:

Remember that in order for the equation of an ellipse to be written in standard form, it must be written in one of the following two ways:

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

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

Where the point (h,k) gives the center of the ellipse, a is half the length of the axis for which it is the denominator, and b is half the length of the axis for which it is the denominator. Looking at our answer choices, we need one that involves addition, has different denominators under the x and y terms, and is equal to 1 on the right side of the equation. We can see that out of all the answer choices, the following is the only one that satisfies those requirements of standard form:

$\frac{(y-3)^2}{4}+\frac{(x+5)^2}{16}=1$

Report an Error

Example Question #4 : Understand Features Of Hyperbolas And Ellipses

What is the equation of an ellipse in standard form with the point (3,6) as its center, an -axis length of 18cm , and a -axis length of 10cm .

Possible Answers:

$\frac{(x-3)^{2}}{81}+\frac{(y-6)^{2}}{25}=1$

$\frac{(x-3)^{2}}{25}+\frac{(y-6)^{2}}{81}=1$

$\frac{(x-3)^{2}}{10}+\frac{(y-6)^{2}}{18}=1$

$\frac{(x-6)^{2}}{81}+\frac{(y-3)^{2}}{25}=1$

$\frac{(x-3)^{2}}{18}+\frac{(y-6)^{2}}{10}=1$

Correct answer:

$\frac{(x-3)^{2}}{81}+\frac{(y-6)^{2}}{25}=1$

Explanation:

The equation of an ellipse in standard form is

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

where (h,k) is the center point, is half the length of its axis in the direction, and is half the length of the axis in the direction.

In this instance, the center point (h,k)= (3,6) , $a=\frac{1}2{(18)}=9$ , and $b=\frac{1}{2}(10)=5$ . Therefore:

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

$\frac{(x-3)^{2}}{9^{2}}+\frac{(y-6)^{2}}{5^{2}}=1$

$\frac{(x-3)^{2}}{81}+\frac{(y-6)^{2}}{25}=1$

Report an Error

Example Question #5 : Understand Features Of Hyperbolas And Ellipses

What is the equation of a circle with radius of and center of (0, 0) ?

Possible Answers:

x^2 + y = 12

5x^2 + 4x + 2 = 10

(x - 2)^2 + (y + 4)^2 = 144

(x - 1)^2 + (y + 1)^2 = 12

x^2 + y^2 = 144

Correct answer:

x^2 + y^2 = 144

Explanation:

Recall that the general equation of a circle is:

(x - h)^2 + (y - k)^2 = r^2 .

The key feature here being that both the and terms are squared.

Thus, we can narrow down the choices without this feature.

Next, consider the center. Since the center here is (0,0) , we should not have any middle terms.

The only choice that fits that description is x^2 + y^2 = 144 .

Report an Error

Example Question #6 : Understand Features Of Hyperbolas And Ellipses

Write the equation for an ellipse with center (1, 4) , foci $(1 \pm \sqrt{14}, 4)$ and a major axis with length 14.

Possible Answers:

$\frac{(x-1)^2 }{ 35 } + \frac{(y-4)^2 }{49 } = 1$

$\frac{(x-1)^2 }{ 196} + \frac{(y-4)^2 }{ 182} = 1$

$\frac{(x-4)^2 }{25 } + \frac{(y-1)^2 }{ 49} = 1$

$\frac{(x+1)^2}{49 } + \frac{(y+4)^2}{35 } =1$

$\frac{(x-1)^2 }{ 49 } + \frac{(y-4)^2 }{35 } = 1$

Correct answer:

$\frac{(x-1)^2 }{ 49 } + \frac{(y-4)^2 }{35 } = 1$

Explanation:

The general equation for an ellipse is $\frac{(x-h)^2 }{a^2 } + \frac{(y-k)^2 }{b^2 } = 1$ , although if we consider a to be half the length of the major axis, a and b might switch depending on if the longer major axis is horizontal or vertical. This general equation has (h,k) as the center, a as the length of half the major axis, and b as the length of half the minor axis.

Because the center of this ellipse is at (1, 4) and the foci are at $(1 \pm \sqrt{14} , 4 )$ , we can see that the foci are $\sqrt{14}$ away from the center, and they are on the horizontal axis. This means that the horizontal axis is the major axis, the one with length 14. Having a length of 14 means that half is 7, so a = 7 . Since the foci are $\sqrt{14}$ away from the center, we know that $c = \sqrt{14}$ . We can solve for b using the equation a^2 - c^2 = b^2 :

$7^2 - \sqrt{14}^2 = b^2$

49 - 14 = b^2

35 = b^2 that's really as far as we need to solve.

Putting all this information into the equation gives:

$\frac{(x-1)^2 }{ 49} + \frac{(y - 4)^2 }{35 } = 1$

Report an Error

Example Question #7 : Understand Features Of Hyperbolas And Ellipses

Find the equation of an ellipse centered at the origin if the major axis is parallel to the x-axis and has a length of units and the minor axis has a length of units.

Possible Answers:

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

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

$\small \small \frac{x^2}{16}+\frac{y^2}{64}=1$

$\small \frac{x^2}{64}+\frac{y^2}{16}=1$

Correct answer:

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

Explanation:

The formula for an ellipse centered at the point (h, k) with a horizontal major axis (ie: parallel to the x-axis) has the formula

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

where $\small a$ and $\small b$ are the lengths of the major and minor axes respectively.

Since the origin is at $\small (0,0)$ and $\small a=8$ and $\small b=4$ in this problem, the equation is

$\small \frac{(x-0)^2}{[\frac{8}{2}]^2}+\frac{(y-0)^2}{[\frac{4}{2}]^2}=1$

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

Report an Error

Example Question #8 : Understand Features Of Hyperbolas And Ellipses

The equation of an ellipse, , is $\frac{(x+3)^2}{9}+\frac{(y-7)^2}{36}=1$ . Which of the following is the correct center and foci of this ellipse?

Possible Answers:

Center= (-3,7) , Foci= (-3,2) and (-3,12)

Center= (3,-7) , Foci= (-2,7) and (8,7)

Center= (-3,7) , Foci= (-8,7) and (2,7)

Center= (3,-7) , Foci= (3,-2) and

Center= (3,6) , Foci= (3,1) and (3,11)

Correct answer:

Center= (-3,7) , Foci= (-3,2) and (-3,12)

Explanation:

Because our equation is already in the format

$\frac{(x-h)^2}{w^2}+\frac{(y-k)^2}{v^2}=1$ ,

we do not have to manipulate the equation. The center of any ellipse in this form will always be . So in this case, our center will be (-3,7) . To find the foci of the ellipse, we must use the equation b^2+c^2=a^2 , where a^2 is the greater of the two denominators in our equation ( w^2 and v^2 ), b^2 is the lesser and is the distance from the center to the foci.

We know that a^2=36 and b^2=9 .

By using a^2-b^2=c^2 , we see that c^2=25 , so c=5 .

We now know that the two foci are going to be units in either direction of the center along the greater axis.

Because the greater denominator is under our term containing , our ellipse will have its greater axis going vertically, rather than horizontally. Therefore, our foci will be units above and below our center, at (-3,2) and (-3,12) .

Report an Error

Example Question #9 : Understand Features Of Hyperbolas And Ellipses

Find the center of the ellipse with the following equation:

$\frac{(x-1)^2}{36}+\frac{(y+2)^2}{9}=1$

Possible Answers:

(-6, -3)

(-1, 2)

(6, 3)

(1, -2)

Correct answer:

(1, -2)

Explanation:

Recall that the standard form of the equation of an ellipse is

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

For the equation given in the question, h=1 and k=-2 .

The center of the ellipse is at (1, -2)

Report an Error

Example Question #10 : Understand Features Of Hyperbolas And Ellipses

Find the center of the ellipse with the following equation:

$\frac{(x+9)^2}{2}+\frac{(y-4)^2}{3}=1$

Possible Answers:

$(-\sqrt2, -\sqrt3)$

(9, -4)

(-9, 4)

$(\sqrt2, \sqrt3)$

Correct answer:

(-9, 4)

Explanation:

Recall that the standard form of the equation of an ellipse is

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

For the equation given in the question, h=-9 and k=4 .

The center of the ellipse is at (-9, 4) .

Report an Error

← Previous 1 2 3 4 5 6 7 8 9 … 12 13 Next →

View Pre-Calculus Tutors

Jesus Ricardo
Certified Tutor

Benemerita Universidad Autónoma de Puebla, Bachelor, Physics.

View Pre-Calculus Tutors

Zara
Certified Tutor

University of Manitoba, Bachelor in Arts, Mathematics.

View Pre-Calculus Tutors

Benjamin
Certified Tutor

Johns Hopkins University, Bachelor in Arts, Mathematics. Johns Hopkins University, Bachelor of Science, Electrical Engineering.

All Precalculus Resources

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

Popular Subjects

Biology Tutors in Phoenix, Biology Tutors in Los Angeles, Statistics Tutors in San Francisco-Bay Area, Math Tutors in Atlanta, Spanish Tutors in Phoenix, Physics Tutors in Seattle, Physics Tutors in Boston, Math Tutors in New York City, ISEE Tutors in San Francisco-Bay Area, SAT Tutors in Miami

Popular Courses & Classes

ACT Courses & Classes in New York City, GRE Courses & Classes in Washington DC, LSAT Courses & Classes in Miami, SAT Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Chicago, Spanish Courses & Classes in Atlanta, GRE Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Atlanta, ISEE Courses & Classes in Phoenix, LSAT Courses & Classes in Chicago

Popular Test Prep

SSAT Test Prep in Phoenix, GRE Test Prep in Phoenix, SSAT Test Prep in Denver, SAT Test Prep in Chicago, SSAT Test Prep in Miami, ACT Test Prep in Seattle, ISEE Test Prep in Boston, ACT Test Prep in New York City, GRE Test Prep in San Diego, ACT Test Prep in Chicago

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 #1 : Understand Features Of Hyperbolas And Ellipses

Example Question #2 : Understand Features Of Hyperbolas And Ellipses

Example Question #3 : Understand Features Of Hyperbolas And Ellipses

Example Question #4 : Understand Features Of Hyperbolas And Ellipses

Example Question #5 : Understand Features Of Hyperbolas And Ellipses

Example Question #6 : Understand Features Of Hyperbolas And Ellipses

Example Question #7 : Understand Features Of Hyperbolas And Ellipses

Example Question #8 : Understand Features Of Hyperbolas And Ellipses

Example Question #9 : Understand Features Of Hyperbolas And Ellipses

Example Question #10 : Understand Features Of Hyperbolas And Ellipses

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: