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Common Core: High School - Geometry : Expressing Geometric Properties with Equations

Study concepts, example questions & explanations for Common Core: High School - Geometry

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All Common Core: High School - Geometry Resources

6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #5 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (4, -10) and (-2, -10) with a major axis distance of .

Possible Answers:

$x^{2} + y^{2} = 1$

$2 x^{2} + 2 y^{2} = 1$

$\left(x - 1\right)^{2} - \frac{1}{8} \left(y + 10\right)^{2} = 1$

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

$\frac{1}{2} \left(x - 4\right)^{2} + \frac{1}{2} \left(y + 10\right)^{2} = 1$

Correct answer:

$\left(x - 1\right)^{2} - \frac{1}{8} \left(y + 10\right)^{2} = 1$

Explanation:

The general equation is

$\frac{1}{b^{2}} \left(- h + y\right)^{2} + \frac{1}{a^{2}} \left(x - 4\right)^{2} = 1$

$\left ( k, \quad h\right )$ is the coordinates of the center of the foci.

is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is.

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 2

Now solve for

a = 1.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = 1 \\y = -10.0$

So the center of the foci is at $\left ( 1, \quad -10\right )$

Now we need to figure out the distance from the center to the foci .

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate

c = -3

Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 1.0 and with -3.

$b^{2} = -8.0$

Now we can substitute these values into the general equation to get.

$\left(x - 1\right)^{2} - \frac{1}{8} \left(y + 10\right)^{2} = 1$

Report an Error

Example Question #6 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (5, 6) and (10, 6) with a major axis distance of .

Possible Answers:

$18 x^{2} + 18 y^{2} = 1$

$\frac{1}{18} \left(x - 5\right)^{2} + \frac{1}{18} \left(y - 6\right)^{2} = 1$

$\frac{1}{81} \left(x - \frac{15}{2}\right)^{2} + \frac{4}{299} \left(y - 6\right)^{2} = 1$

$x^{2} + y^{2} = 1$

$\frac{x^{2}}{18} + \frac{y^{2}}{18} = 1$

Correct answer:

$\frac{1}{81} \left(x - \frac{15}{2}\right)^{2} + \frac{4}{299} \left(y - 6\right)^{2} = 1$

Explanation:

The general equation is

$\frac{1}{b^{2}} \left(- h + y\right)^{2} + \frac{1}{a^{2}} \left(x - 5\right)^{2} = 1$

$\left ( k, \quad h\right )$ is the coordinates of the center of the foci.

is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is.

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 18

Now solve for

a = 9.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = \frac{15}{2} \\\\ y = 6.0$

So the center of the foci is at $\left ( \frac{15}{2}, \quad 6\right )$

Now we need to figure out what the distance from the center to where the foci is.

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate

$c = \frac{5}{2}$

Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 9.0 and with $\frac{5}{2}$

$b^{2} = 74.75$

Now we can substitute these values into the general equation to get.

$\frac{1}{81} \left(x - \frac{15}{2}\right)^{2} + \frac{4}{299} \left(y - 6\right)^{2} = 1$

Report an Error

Example Question #7 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (-10, -10) and (4, -10) with a major axis distance of .

Possible Answers:

$\frac{1}{36} \left(x + 3\right)^{2} - \frac{1}{13} \left(y + 10\right)^{2} = 1$

$12 x^{2} + 12 y^{2} = 1$

$x^{2} + y^{2} = 1$

$\frac{1}{12} \left(x + 10\right)^{2} + \frac{1}{12} \left(y + 10\right)^{2} = 1$

$\frac{x^{2}}{12} + \frac{y^{2}}{12} = 1$

Correct answer:

$\frac{1}{36} \left(x + 3\right)^{2} - \frac{1}{13} \left(y + 10\right)^{2} = 1$

Explanation:

The general equation is

$\frac{1}{b^{2}} \left(- h + y\right)^{2} + \frac{1}{a^{2}} \left(x + 10\right)^{2} = 1$

$\left ( k, \quad h\right )$ is the coordinates of the center of the foci.

is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is.

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 12

Now solve for

a = 6.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = -3 \\y = -10.0$

So the center of the foci is at $\left ( -3, \quad -10\right )$

Now we need to figure out what the distance from the center to the foci is.

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate

c = 7
Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 6.0 and with 7

$b^{2} = -13.0$

Now we can substitute these values into the general equation to get.

$\frac{1}{36} \left(x + 3\right)^{2} - \frac{1}{13} \left(y + 10\right)^{2} = 1$

Report an Error

Example Question #8 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (5, -1) and (-6, -1) with a major axis distance of .

Possible Answers:

$14 x^{2} + 14 y^{2} = 1$

$\frac{1}{49} \left(x + \frac{1}{2}\right)^{2} + \frac{4}{75} \left(y + 1\right)^{2} = 1$

$\frac{1}{14} \left(x - 5\right)^{2} + \frac{1}{14} \left(y + 1\right)^{2} = 1$

$\frac{x^{2}}{14} + \frac{y^{2}}{14} = 1$

$x^{2} + y^{2} = 1$

Correct answer:

$\frac{1}{49} \left(x + \frac{1}{2}\right)^{2} + \frac{4}{75} \left(y + 1\right)^{2} = 1$

Explanation:

The general equation is

$\frac{1}{b^{2}} \left(- h + y\right)^{2} + \frac{1}{a^{2}} \left(x - 5\right)^{2} = 1$

$\left ( k, \quad h\right )$ is the coordinates of the center of the foci.

is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is ().

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 14

Now solve for

a = 7.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = - \frac{1}{2} \\\\y = -1.0$

So the center of the foci is at $\left ( - \frac{1}{2}, \quad -1\right )$

Now we need to figure out what the distance from the center to the foci .

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate

$c = - \frac{11}{2}$

Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 7.0 and with $- \frac{11}{2}$

$b^{2} = 18.75$

Now we can substitute these values into the general equation to get.

$\frac{1}{49} \left(x + \frac{1}{2}\right)^{2} + \frac{4}{75} \left(y + 1\right)^{2} = 1$

Report an Error

Example Question #9 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (-8, -5 ) and (3, -5) with a major axis distance of .

Possible Answers:

$\frac{x^{2}}{18} + \frac{y^{2}}{18} = 1$

$18 x^{2} + 18 y^{2} = 1$

$\frac{1}{18} \left(x + 8\right)^{2} + \frac{1}{18} \left(y + 5\right)^{2} = 1$

$\frac{1}{81} \left(x + \frac{5}{2}\right)^{2} + \frac{4}{203} \left(y + 5\right)^{2} = 1$

$x^{2} + y^{2} = 1$

Correct answer:

$\frac{1}{81} \left(x + \frac{5}{2}\right)^{2} + \frac{4}{203} \left(y + 5\right)^{2} = 1$

Explanation:

The general equation is

$\frac{1}{b^{2}} \left(- h + y\right)^{2} + \frac{1}{a^{2}} \left(x + 8\right)^{2} = 1$

$\left ( k, \quad h\right )$ is the coordinates of the center of the foci.

is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is ().

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 18

Now solve for

a = 9.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = - \frac{5}{2} \\\\y = -5.0$

So the center of the foci is at $\left ( - \frac{5}{2}, \quad -5\right )$

Now we need to figure out what the distance from the center to the foci is ().

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate

$c = \frac{11}{2}$
Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 9.0 and with $\frac{11}{2}$

$b^{2} = 50.75$

Now we can substitute these values into the general equation to get.

$\frac{1}{81} \left(x + \frac{5}{2}\right)^{2} + \frac{4}{203} \left(y + 5\right)^{2} = 1$

Report an Error

Example Question #10 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (7, -1) and (0, -1) with a major axis distance of .

Possible Answers:

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

$\frac{1}{64} \left(x - \frac{7}{2}\right)^{2} + \frac{4}{207} \left(y + 1\right)^{2} = 1$

$x^{2} + y^{2} = 1$

$16 x^{2} + 16 y^{2} = 1$

$\frac{1}{16} \left(x - 7\right)^{2} + \frac{1}{16} \left(y + 1\right)^{2} = 1$

Correct answer:

$\frac{1}{64} \left(x - \frac{7}{2}\right)^{2} + \frac{4}{207} \left(y + 1\right)^{2} = 1$

Explanation:

The general equation is

$\frac{1}{b^{2}} \left(- h + y\right)^{2} + \frac{1}{a^{2}} \left(x - 7\right)^{2} = 1$

$\left ( k, \quad h\right )$ is the coordinates of the center of the foci.

is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is ( $\uptext{a}$ ).

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 16

Now solve for

a = 8.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = \frac{7}{2} \\\\y = -1.0$

So the center of the foci is at $\left ( \frac{7}{2}, \quad -1\right )$

Now we need to figure out what the distance from the center to the foci is ( $\uptext{c}$ ).

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate

$c = - \frac{7}{2}$
Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 8.0 and with $- \frac{7}{2}$

$b^{2} = 51.75$

Now we can substitute these values into the general equation to get.

$\frac{1}{64} \left(x - \frac{7}{2}\right)^{2} + \frac{4}{207} \left(y + 1\right)^{2} = 1$

Report an Error

Example Question #31 : Expressing Geometric Properties With Equations

Find the equation of an ellipse that has foci at (-6, 9) and (7, 9) with a major axis distance of .

Possible Answers:

$16 x^{2} + 16 y^{2} = 1$

$\frac{1}{16} \left(x + 6\right)^{2} + \frac{1}{16} \left(y - 9\right)^{2} = 1$

$x^{2} + y^{2} = 1$

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

$\frac{1}{64} \left(x - \frac{1}{2}\right)^{2} + \frac{4}{87} \left(y - 9\right)^{2} = 1$

Correct answer:

$\frac{1}{64} \left(x - \frac{1}{2}\right)^{2} + \frac{4}{87} \left(y - 9\right)^{2} = 1$

Explanation:

The general equation is

$\frac{1}{b^{2}} \left(- h + y\right)^{2} + \frac{1}{a^{2}} \left(x + 6\right)^{2} = 1$

$\left ( k, \quad h\right )$ is the coordinates of the center of the foci.

is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is ().

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 16

Now solve for

a = 8.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = \frac{1}{2} \\\\y = 9.0$

So the center of the foci is at $\left ( \frac{1}{2}, \quad 9\right )$

Now we need to figure out what the distance from the center to the foci is ().

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate

$c = \frac{13}{2}$
Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 8.0 and with $\frac{13}{2}$

$b^{2} = 21.75$

Now we can substitute these values into the general equation to get.

$\frac{1}{64} \left(x - \frac{1}{2}\right)^{2} + \frac{4}{87} \left(y - 9\right)^{2} = 1$

Report an Error

Example Question #32 : Expressing Geometric Properties With Equations

Find the equation of an ellipse that has foci at (-4, -6) and (-9, -6) with a major axis distance of .

Possible Answers:

$x^{2} + y^{2} = 1$

$\left(x + \frac{13}{2}\right)^{2} - \frac{4}{21} \left(y + 6\right)^{2} = 1$

$2 x^{2} + 2 y^{2} = 1$

$\frac{1}{2} \left(x + 4\right)^{2} + \frac{1}{2} \left(y + 6\right)^{2} = 1$

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

Correct answer:

$\left(x + \frac{13}{2}\right)^{2} - \frac{4}{21} \left(y + 6\right)^{2} = 1$

Explanation:

The general equation is

$\frac{1}{b^{2}} \left(- h + y\right)^{2} + \frac{1}{a^{2}} \left(x + 4\right)^{2} = 1$

$\left ( k, \quad h\right )$ is the coordinates of the center of the foci.

is the distance to the foci on the x-axis from the center, and is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is ().

Since we know the distance of the major axis, all we need to do is set up a simple equation.

2 a = 2

Now solve for

a = 1.0

The next step is to find the center between the foci.

To find the center, we simply find the average between the coordinates.

$\\x = - \frac{13}{2} \\\\y = -6.0$

So the center of the foci is at $\left ( - \frac{13}{2}, \quad -6\right )$

Now we need to figure out what the distance from the center to the foci is ().

We do this by taking the x-coordinate from one of the foci and subtracting it from the center x-coordinate

$c = - \frac{5}{2}$

Now the last part is to find .

We can find it by using the following equation.

$b^{2} = a^{2} - c^{2}$

We simply substitute with 1.0 and with $- \frac{5}{2}$

$b^{2} = -5.25$

Now we can substitute these values into the general equation to get.

$\left(x + \frac{13}{2}\right)^{2} - \frac{4}{21} \left(y + 6\right)^{2} = 1$

Report an Error

Example Question #1 : Prove Simple Geometric Theorems By Using Coordinates: Ccss.Math.Content.Hsg Gpe.B.4

True or False:

The point (2, 2) lies on the ellipse

$\frac{1}{15}$ $\left(x - 2\right)^{2} + \frac{1}{11} \left(y + 1\right)^{2} = 1$

Possible Answers:

True

False

Correct answer:

False

Explanation:

In order to solve this problem, we need to plug in the point, and see if both sides of the equation are equal.

$\frac{1}{15}$ $\left(x - 2\right)^{2} + \frac{1}{11} \left(y + 1\right)^{2} = 1$

$\frac{1}{15}$ $\left(2 - 2\right)^{2} + \frac{1}{11} \left(2 + 1\right)^{2} = 1$

$\frac{1}{15}$ $\cdot 0 + \frac{1}{11} \cdot 9= 1$

$\frac{9}{11} = 1$

The point does not lie on the ellipse, because each side is not equal to each other. So our answer is False.

Report an Error

Example Question #2 : Prove Simple Geometric Theorems By Using Coordinates: Ccss.Math.Content.Hsg Gpe.B.4

True or False:

The point (3, 2) lies on the ellipse

$\frac{1}{9} \left(x - 4\right)^{2} + \frac{1}{12} \left(y + 1\right)^{2} = 1$

Possible Answers:

False

True

Correct answer:

False

Explanation:

In order to solve this problem, we need to plug in the point, and see if both sides of the equation are equal.

$\frac{1}{9} \left(x - 4\right)^{2} + \frac{1}{12} \left(y + 1\right)^{2} = 1$

$\frac{1}{9} \left(3 - 4\right)^{2} + \frac{1}{12} \left(2 + 1\right)^{2} = 1$

$\frac{1}{9} \left(-1\right)^{2} + \frac{1}{12} \left(3\right)^{2} = 1$

$\frac{1}{9} + \frac{1}{12} \cdot 9 = 1$

$\frac{1}{9} + \frac{9}{12} = 1$

$\frac{31}{36} = 1$

The point does not lie on the ellipse, because each side is not equal to each other. So our answer is False.

Report an Error

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All Common Core: High School - Geometry Resources

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Common Core: High School - Geometry : Expressing Geometric Properties with Equations

Study concepts, example questions & explanations for Common Core: High School - Geometry

All Common Core: High School - Geometry Resources

Example Questions

Example Question #5 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Example Question #6 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Example Question #7 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Example Question #8 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Example Question #9 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Example Question #10 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Example Question #31 : Expressing Geometric Properties With Equations

Example Question #32 : Expressing Geometric Properties With Equations

Example Question #1 : Prove Simple Geometric Theorems By Using Coordinates: Ccss.Math.Content.Hsg Gpe.B.4

Example Question #2 : Prove Simple Geometric Theorems By Using Coordinates: Ccss.Math.Content.Hsg Gpe.B.4

All Common Core: High School - Geometry Resources

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