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

← Previous 1 2 3 4 5 6 7 8 9 … 26 27 Next →

Example Question #1 : Circles

What is the center and radius of the circle indicated by the equation?

(x-2)^2+y^2=36

Possible Answers:

$(2,0),\ r=6$

$(-2,0),\ r=6$

$(2,0),\ r=36$

$(-2,0),\ r=36$

Correct answer:

$(2,0),\ r=6$

Explanation:

A circle is defined by an equation in the format (x-h)^2+(y-k)^2=r^2 .

The center is indicated by the point (h,k) and the radius .

In the equation (x-2)^2+(y)^2=36=6^2 , the center is (2,0) and the radius is .

Report an Error

Example Question #1 : Conic Sections

What is the equation of the circle with center at (4, -5) and a point on the circle at (4, -2)? What is the radius?

Possible Answers:

$\left ( x-4\right )^2 + (y-5)^2 = 9$

r = 9

$\left ( x+4\right )^2 + (y+5)^2 = 9$

r = 3

$\left ( x-4\right )^2 + (y+5)^2 = 9$

r = 3

$\left ( x-4\right )^2 + (y+5)^2 = 3$

r = 3

(x+4)^2-(y+5)^2 = 3

r = 9

Correct answer:

$\left ( x-4\right )^2 + (y+5)^2 = 9$

r = 3

Explanation:

The form most often used for circles is the following general equation:

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

where (h, k) are the coordinates of the center and r is the radius. We are given the coordinates of the center as (4, -5), so h is 4 and k is -5.

(x-4)^2 +(y-(-5)))^2 = r^2

(x-4)^2 +(y+5)^2 = r^2

We still need to find the radius. We can do this by plugging in the second given point, (4, -2).

(4 - 4)^2 + (-2 + 5)^2 = r^2

0^2 + 3^2 = r^2

0 + 9 = r^2

9 = r^2

Hence the formula of the circle is

(x-4)^2 +(y+5)^2 = 9 ,

and the radius is r = 3 .

Report an Error

Example Question #1 : Circles

Find the center and radius of this circle:

x^2+y^2-8x+2y-19=0

Possible Answers:

c=(4,-1)

$r=\sqrt{2}$

c=(8,-2)

r=6

c=(4,-1)

r=6

c=(4,-1)

$r=\sqrt{19}$

c=(-4,1)

r=6

Correct answer:

c=(4,-1)

r=6

Explanation:

To solve this problem we need to get our function into the form:

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

where center=(h,k) and r=radius. To do this we need to complete the square with both the x an y parts of the equation

(x^2-8x+16)+(y^2+2x+1)=19+16+1

then we factor our trinomials and add our constants and we get:

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

so our center is (4,-1) and our radius is 6

Report an Error

Example Question #1 : Conic Sections

Which function will graph a circle of radius , centered at (-2,8) ?

Possible Answers:

(x+2)^2+(y-8)^2=4

(x-2)^2+(y-8)^2=16

(x+2)^2+(y-8)^2 = 16

(x+2)^2-(y-8)^2=16

(x-2)^2+(y+8)^2=16

Correct answer:

(x+2)^2+(y-8)^2 = 16

Explanation:

The general form for the function of a circle of radius r and centered at (a,b) is:

(x-a)^2+(y-b)^2=r^2 .

The circle is centered at (-2,8) , therefore we substitute in for and for . The radius is so r^2 becomes .

The final form should be:

(x+2)^2+(y-8)^2=16 .

Report an Error

Example Question #1 : Conic Sections

Find the focal points of the conic below:

$27x^{^{2}}+16y^{^{2}}+324x-96y+684 = 0$

Possible Answers:

$\left ( -6,14\right ) , \left (-6,-8 \right )$

$\left ( -6,3+\sqrt{11}\right ) , \left (-6,3-\sqrt{11} \right )$

$\left ( 6,-3+\sqrt{11}\right ) , \left (6,-3-\sqrt{11} \right )$

$\left ( -6+\sqrt{11},3\right ),\left ( -6-\sqrt{11},3\right )$

$\left ( -6,3+\sqrt{13}\right ) , \left (-6,3-\sqrt{13} \right )$

Correct answer:

$\left ( -6,3+\sqrt{11}\right ) , \left (-6,3-\sqrt{11} \right )$

Explanation:

The first thing we want to do is put the conic (an ellipse because the x²and the y² terms have the same sign) into a better form i.e.

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

where (h,k) is the center of our ellipse.

We will continue by completing the square for both the x and y binomials.

First we seperate them into two trinomials:

$\left ( 27x^2+324x+0\right )+\left ( 16y^2-96y+0\right )=-684$

then we pull a 27 out of the first one and a 16 out of the second

$27\left ( x^2+12x+0\right )+16\left ( y^2-6y+0\right )=-684$

then we add the correct constant to each trinomial (being sure to add the same amount to the other side of our equation.

$27\left ( x^2+12x+36\right )+16\left ( y^2-6y+9\right )=-684+27(36)+16(9)$

then we factor our trinomials and divide by 16 and 27 to get

$\frac{\left ( x+6\right )^2}{16}+\frac{\left ( y-3\right )^2}{27}=1$

so the center of our ellipse is (-6,3) and we calculate the distance from the focal points to the center by the equation:

$c^2=\left | a^2-b^2\right |$

and we know that our ellipse is stretched in the y direction because b>a so our focal points will be c displaced from our center.

with $c=\sqrt{11}$

our focal points are $\left ( -6,3+\sqrt{11}\right ) , \left (-6,3-\sqrt{11} \right )$

Report an Error

Example Question #1 : Pre Calculus

Find the center of this ellipse:

$33x^2+13y^2-396x+26\sqrt{7}y+850=0$

Possible Answers:

(-23,17)

$(6,-\sqrt7)$

(36,7)

$(6,\sqrt7)$

(23,-17)

Correct answer:

$(6,-\sqrt7)$

Explanation:

To find the center of this ellipse we need to put it into a better form. We do this by rearranging our terms and completing the square for both our y and x terms.

$33(x^2-12x)+13(y^2+2\sqrt7y)=-850$

completing the square for both gives us this.

$33(x^2-12x+36)+13(y^2+2\sqrt7y+7)=429$

$33(x-6)^2+13(y+\sqrt7)^2=429$

we could divide by 429 but we have the information we need. The center of our ellipse is $(6,-\sqrt7 )$

Report an Error

Example Question #7 : Conic Sections

What is the equation of the elipse centered at the origin and passing through the point (5, 0) with major radius 5 and minor radius 3?

Possible Answers:

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

$\frac{x^2}{5} + \frac{y^2}{3} = 1$

$\frac{x^2}{3} + \frac{y^2}{5} = 1$

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

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

Correct answer:

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

Explanation:

The equation of an ellipse is

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

where a is the horizontal radius, b is the vertical radius, and (h, k) is the center of the ellipse. In this case we are told that the center is at the origin, or (0,0), so both h and k equal 0. That brings us to:

$\frac{(x-0)^2}{a^2} + \frac{(y-0)^2}{b^2} = 1$

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

We are told about the major and minor radiuses, but the problem does not specify which one is horizontal and which one vertical. However it does tell us that the ellipse passes through the point (5, 0), which is in a horizontal line with the center, (0, 0). Therefore the horizontal radius is 5.

The vertical radius must then be 3. We can now plug these in:

$\frac{x^2}{5^2} + \frac{y^2}{3^2} = 1$

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

Report an Error

Example Question #1 : Ellipses

An ellipse is centered at (-3, 2) and passes through the points (-3, 6) and (4, 2). Determine the equation of this eclipse.

Possible Answers:

$\frac{(x+3)^2}{16} + \frac{(y-2)^2}{49} = 1$

$\frac{(x-3)^2}{49} + \frac{(y-2)^2}{16} = 1$

$\frac{(x+3)^2}{49} + \frac{(y-2)^2}{16} = 1$

$\frac{(x-3)^2}{49} + \frac{(y+2)^2}{16} = 1$

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

Correct answer:

$\frac{(x+3)^2}{49} + \frac{(y-2)^2}{16} = 1$

Explanation:

The usual form for an ellipse is

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

where (h, k) is the center of the ellipse, a is the horizontal radius, and b is the vertical radius.

Plug in the coordinate pair:

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

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

Now we have to find the horizontal radius and the vertical radius. Let's compare points; we are told the ellipse passes through the point (-3, 6), which is vertically aligned with the center. Therefore the vertical radius is 4.

Similarly, the ellipse passes through the point (4, 2), which is horizontally aligned with the center. This means the horizontal radius must be 7.

Substitute:

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

$\frac{(x+3)^2}{49} + \frac{(y-2)^2}{16} = 1$

Report an Error

Example Question #1 : Conic Sections

What is the shape of the graph indicated by the equation?

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

Possible Answers:

Parabola

Ellipse

Circle

Hyperbola

Correct answer:

Ellipse

Explanation:

An ellipse has an equation that can be written in the format $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$ . The center is indicated by (h,k) , or in this case (0,0) .

Report an Error

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

← Previous 1 2 3 4 5 6 7 8 9 … 26 27 Next →

View Pre-Calculus Tutors

Magdy
Certified Tutor

Cairo University, Egypt, Bachelor of Science, Electrical Engineering. New Mexico State University-Main Campus, Doctor of Phil...

View Pre-Calculus Tutors

Midori
Certified Tutor

University of Connecticut, Bachelor in Arts, Applied Mathematics. University of Southern California, Master of Arts Teaching,...

View Pre-Calculus Tutors

Jesus Ricardo
Certified Tutor

Benemerita Universidad Autónoma de Puebla, Bachelor, Physics.

All Precalculus Resources

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

Popular Subjects

Spanish Tutors in Chicago, Statistics Tutors in Miami, ACT Tutors in Dallas Fort Worth, ACT Tutors in Philadelphia, SAT Tutors in Philadelphia, Chemistry Tutors in Seattle, ACT Tutors in Seattle, LSAT Tutors in Chicago, LSAT Tutors in San Francisco-Bay Area, Algebra Tutors in Los Angeles

Popular Courses & Classes

ACT Courses & Classes in Philadelphia, ISEE Courses & Classes in Chicago, ISEE Courses & Classes in Philadelphia, ISEE Courses & Classes in Houston, SSAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Seattle, ISEE Courses & Classes in Washington DC, LSAT Courses & Classes in Washington DC, Spanish Courses & Classes in New York City, SSAT Courses & Classes in New York City

Popular Test Prep

ISEE Test Prep in Chicago, ISEE Test Prep in Seattle, SAT Test Prep in Miami, LSAT Test Prep in Chicago, LSAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in San Francisco-Bay Area, ISEE Test Prep in Philadelphia, GRE Test Prep in Chicago, SAT Test Prep in Denver, LSAT Test Prep in Philadelphia

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 #1 : Circles

Example Question #1 : Conic Sections

Example Question #1 : Circles

Example Question #1 : Conic Sections

Example Question #1 : Conic Sections

Example Question #1 : Pre Calculus

Example Question #7 : Conic Sections

Example Question #1 : Ellipses

Example Question #1 : Conic Sections

Example Question #10 : Conic Sections

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: