Geometry Worksheets

Practice and master geometry concepts with helpful walkthroughs from our elite math educators, then try practice math problems to sharpen your skills.

Conic Sections

By Alison Rodriguez

Conic sections are formed when a plane intersects a cone.What results is either a circle, ellipse, parabola, or hyperbola. This article will review each of the conic sections, what they are, what they look like, and how to use them.

Why is this concept useful?

Conics are used in architecture, physics, optics, astronomy, navigation, and many other real-word applications.

Where does this concept fit into the curriculum?

Geometry, Pre-calculus and Calculus

How can we use conic sections?

Let’s begin with a visual of what the conic sections look like:

A circle is formed when the plane is perpendicular to the axis and goes through one of the cones.

An ellipse is formed when the plane intersects one of the cones but is not perpendicular to the axis.

A parabola is formed when the plane intersects one of the cones and is not perpendicular to the axis

A hyperbola is formed when a plane intersects both of the cones.

Depending on where the plane intersects the cone, it will change the size and shape of the conic section. All conic sections have the same general form:

Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^{2} +Bxy+Cy^{2} +Dx+Ey+F=0

Where A, B, C, D, E, and F are all real numbers.

The following will be a summary of each type of conic section.

1. Circles

A circle is the set of all points in a plane that are a distance r from a given point, called the center. The distance r is the radius of the circle.

2. Ellipse

An ellipse is a set of all points (x, y) in a plane, the sum of whose distances from two fixed points called foci is constant.

The following are important values to be aware of for the ellipse.

C = Center (h, k)

F = Foci (two fixed points)

V = Vertices, found on the major axis

CoV = co-vertices, found on the minor axis

c = distance from the center to the foci

a = distance from the center to the vertices

b = distance from the center to covertices

e = Eccentricity of the ellipse, found by doing

ca\frac{c}{a}

(the closer to zero, the more of a circle shape, the closer to one the more elongated)

Major axis length = 2a

Minor axis length = 2b

The following image shows a horizontal ellipse, centered at the origin, whose major axis is horizontal and minor is vertical.

The equation for a horizontal ellipse in standard form is:

(x−h)2a3+(y−k)2b2=1\frac{( x-h)^{2}}{a^{3}} +\frac{( y-k)^{2}}{b^{2}} =1

The following image shows a vertical ellipse, centered at the origin, whose major axis is vertical and minor is horizontal:

The equation for a vertical ellipse is:

(x−h)2b2+(y−h)2a2=1\frac{( x-h)^{2}}{b^{2}} +\frac{( y-h)^{2}}{a^{2}} =1

Note that b and a switch places for either horizontal or vertical ellipses.

One other important thing to find would be c. The formula to find either a, b, or c looks like the Pythagorean theorem, but does have a difference in where the letters go. This formula can help find

a2=b2+c2 or c2=a2−b2a^{2} =b^{2} +c^{2} \ or\ c^{2} =a^{2} -b^{2}

3. Hyperbola

An ellipse is a set of all points (x, y) in a plane, the difference of whose distances from two fixed points called foci is constant. Pictured below:

Hyperbolas have a a major and minor axis known as the transverse and conjugate axes respectively.

The following a horizontal hyperbola, centered at the origin, whose transverse axis is horizontal and conjugate is vertical

The standard form of a horizontal hyperbola is:

(y−h)2a2−(x−k)2b2=1\frac{( y-h)^{2}}{a^{2}} -\frac{( x-k)^{2}}{b^{2}} =1

Note how the x and y switch places, not a and b like in an ellipse. Additionally, there is subtraction between the fractions.

4. Parabola

A parabola can be defined differently than what you may be used to! We have this new line in, called the directrix, such that any point on the parabola is equidistant to the focus and the directrix. See the picture below!

The standard form of a parabola takes on many forms to determine the direction in which it will open up.

(x−h)2=4p(y−k)( x-h)^{2} =4p( y-k)

has a vertex at (h,k) and will open upwards.

(x−h)2=−4p(y−k)( x-h)^{2} =-4p( y-k)

has a vertex at (h,k) and will open upwards.

(y−k)2=4p(x−h)( y-k)^{2} =4p( x-h)

has a vertex at (h,k) and will open to the right.

(y−k)2=−4p(x−h)( y-k)^{2} =-4p( x-h)

has a vertex at (h,k) and will open to the left. There also exists a segment, known as the latus rectum (L.R.) which goes through the focus and is perpendicular to the axis of symmetry and has endpoints on the parabola. The length of the L.R. is |4p| units where p is the distance from the vertex to the focus. See the picture below!

Sample Math Problems

1. Rewrite the following equation in standard form. State the center and radius of the circle.

x2+8x+y2−2y−64=0x^{2} +8x+y^{2} -2y-64=0

Solution: The standard form of a circle is

(x−h)2+(y−k)2=r2( x-h)^{2} +( y-k)^{2} =r^{2}

So for this problem, we will need to complete the square in order to factor. Start by moving the constant at the end to the right side of the equal sign.

x2+8x+ + y2−2y +  =64x^{2} +8x+\ +\ y^{2} -2y\ +\ \ =64

Then, fill in the blanks by completing the square

(82)2=16\left(\frac{8}{2}\right)^{2} =16

and

(−21)2=1\left(\frac{-2}{1}\right)^{2} =1
x2+8x+16+y2−2y+1=64+16+1x^{2} +8x+16+y^{2} -2y+1=64+16+1
(x+4)2+(y−1)2=81 \begin{array}{l} ( x+4)^{2} +( y-1)^{2} =81\\ \end{array}

This is a circle whose center is at (-4, 1) and radius is equal to 9.

2. Identify the center, vertices, co-vertices, and foci

(x−1)29+(y+5)24=1\frac{( x-1)^{2}}{9} +\frac{( y+5)^{2}}{4} =1

Solution: This is a horizontal ellipse as the bigger number is underneath the x, therefore a = 3 and b = 2.

Use

a2=b2+c2a^{2} =b^{2} +c^{2}

to find c.

9=4+c29=4+c^{2}
5=c\sqrt{5} =c

Center: This is found from the original equation, the center is (1, -5)

Vertices: Since this is a horizontal ellipse, to find the vertices you will add/subtract “a” to the x-value of the center

(1 + 3, -5) → (4,-5)

(1 - 3, -5) → (-2, -5)

Co-vertices: Since this is a horizontal ellipse, to find the co-vertices you will add/subtract “b” to the y-value of the center.

(1, -5 + 2) → (-1, -3)

(1, -5 - 2) → (-1, -7)

Foci: Since this is a horizontal ellipse, to find the foci you will add/subtract the value of c that was found to the x-value.

(1+5,−5)\left( 1+\sqrt{5} ,-5\right)
(1−5,−5)\left( 1-\sqrt{5} ,-5\right)

Knowing all of this information can allow you to sketch the graph of this function.



3. Write the equation of a hyperbola with vertices at (4, -1) and (-6, -1) and foci (8, -1) and (-10, -1)

Solution:

Using the vertices, we can find the midpoint to find the center of this hyperbola.

(4+(−6)2,−1+(−1)2) → (−1,−1)\left(\frac{4+( -6)}{2} ,\frac{-1+( -1)}{2}\right) \ \rightarrow \ ( -1,-1)

The value of a is the distance from the center to a vertex.

From (-1, -1) to (4, -1) the distance is 5, so a = 5

The value of c is the distance from the center to the foci.

From (-1, -1) to (8, -1) the distance is 9, so c = 9

To find b, use

c2=a2+b2c^{2} =a^{2} +b^{2}
92=52+b29^{2} =5^{2} +b^{2}
81=25+b281=25+b^{2}
56=b256=b^{2}

Since the values line up horizontally, this is a horizontal hyperbola meaning x is listed first and

a2a^{2}

is underneath it. The final equation is:

(x+1)225−(y+1)256=1\frac{( x+1)^{2}}{25} -\frac{( y+1)^{2}}{56} =1

4. A domed ceiling is a parabolic surface. Ten meters down from the top of the dome, the ceiling is 15 meters wide. For the best lighting on the floor, a light source should be placed at the focus of the parabolic surface. What is the equation to represent the ceiling?

Solution:

The ceiling would look like a parabola that is facing downwards, so we will be using the equations

(x−h)2=−4p(y−k)2( x-h)^{2} =-4p( y-k)^{2}

It does not specify where the vertex is, so we will place it at the origin of (0, 0). Therefore the equation can simplify to

x2=−4py2x^{2} =-4py^{2}

To find the value of p, we know that another set of points, 10 meters down from the vertex, split along the axis of symmetry 7.5 meters on each side for a total of 15 meters. We can identify these points as (-7.5, -10) and (7.5, -10).

(7.5)2=−4p(−10)( 7.5)^{2} =-4p( -10)
56.25=40p56.25=40p

1.4=p, therefore 4p = 5.6 The final equation is

x2=−5.6yx^{2} =-5.6y

Download FREE Math Resources

Take advantage of our free downloadable resources and study materials for at-home learning.

8 Math Hacks and Tricks to Turn Your ‘Okay’ Math Student Into a Math Champion!

One thing we teach our students at Thinkster is that there are multiple ways to solve a math problem. This helps our students learn to think flexibly and non-linearly.

Get PDF

How to Make Sure Your Child is Highly Successful and Becomes a Millionaire

As a parent, you hope your child is extremely successful and likely become the next Gates, Zuckerberg, or Meg Whitman. To set your child on the right path, there are many skills and traits that you can start building and nurturing now. Doing so plants the seeds for future success.

Get PDF

Practice Math Problems

1. Rewrite the following equation in standard form. State the center and radius of the circle.

x2−5x+y2+6y+7=0x^{2} -5x+y^{2} +6y+7=0

2. Identify the center, vertices, co-vertices, and foci

(x+2)216+(y−3)225=1\frac{( x+2)^{2}}{16} +\frac{( y-3)^{2}}{25} =1

3. Find the standard from of the equation of the hyperbola with vertices at (0, 6) and (0, -6) and foci at (0, 7) and (0, -7)

4. A car headlight mirror has a parabolic cross section with diameter of 6 in and a depth of 5 inches. Write an equation to represent the headlight.

Related Topics

By Puja Varude

Normal Distribution

By Patricia Martin

Tossing a Coin

By Mandi Elam

Properties of Exponents

By Debi DalPezzo

Solve for x

Math Tutoring to Boost Your Child’s Math Skills & Scores by 90% in Just 3 Months – Guaranteed!

Does your child struggle with math homework or understanding tricky math concepts? Do they do okay in math, but express excitement to learn new material or advanced math?

A Thinkster Math tutor provides one-to-one support to help elementary, middle school, and high school students build confidence and master math subjects like K-8 math, pre-algebra, algebra, geometry, calculus, and more.

Our expert math tutors customize math lessons to your child’s unique needs, making learning math fun and effective. We help students improve grades, develop strong critical thinking skills through solving word problems, excel in standardized tests, and develop strong problem-solving skills.

Our expert math tutors are ready to help make your child a champion and develop strong math mastery! Sign up for our 7-day free trial and get the best math tutor for your child today!

Start 7-Day Free Trial