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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 #9 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( 6, 8\right ) \\ \text{directrix } y = 10$

Possible Answers:

$\left(x - 2\right)^{2}$

$y = \frac{x}{16} \left(- x + 12\right)$

$\left(x - 8\right)^{2} + \left(x - 6\right)^{2}$

$y = \frac{x}{8} \left(- x + 12\right)$

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

Correct answer:

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

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute 6 for a 8 for b and 10 for y

$y_{0}^{2} - 20 y_{0} + 100 = x_{0}^{2} - 12 x_{0} + y_{0}^{2} - 16 y_{0} + 100$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}}{4} \left(- x_{0} + 12\right)$

So our answer is then

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

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Example Question #10 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( -10, -3\right ) \\\text{directrix } y = -4$

Possible Answers:

$y = \frac{x^{2}}{4} + 5 x + \frac{93}{4}$

$y = \frac{x^{2}}{8} + \frac{5 x}{2} + \frac{93}{8}$

$\left(x + 3\right)^{2} + \left(x + 10\right)^{2}$

$\left(x - 2\right)^{2}$

$y = \frac{x^{2}}{2} + 10 x + \frac{93}{2}$

Correct answer:

$y = \frac{x^{2}}{2} + 10 x + \frac{93}{2}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -10 for a -3 for b and -4 for y

$y_{0}^{2} + 8 y_{0} + 16 = x_{0}^{2} + 20 x_{0} + y_{0}^{2} + 6 y_{0} + 109$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{2} + 10 x_{0} + \frac{93}{2}$

So our answer is then

$y = \frac{x^{2}}{2} + 10 x + \frac{93}{2}$

Report an Error

Example Question #11 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( 2, 5\right ) \\\text{directrix } y = -6$

Possible Answers:

$\left(x - 2\right)^{2}$

$y = \frac{x^{2}}{88} - \frac{x}{22} - \frac{7}{88}$

$\left(x - 5\right)^{2} + \left(x - 2\right)^{2}$

$y = \frac{x^{2}}{22} - \frac{2 x}{11} - \frac{7}{22}$

$y = \frac{x^{2}}{44} - \frac{x}{11} - \frac{7}{44}$

Correct answer:

$y = \frac{x^{2}}{22} - \frac{2 x}{11} - \frac{7}{22}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute 2 for a 5 for b and -6 for y

$y_{0}^{2} + 12 y_{0} + 36 = x_{0}^{2} - 4 x_{0} + y_{0}^{2} - 10 y_{0} + 29$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{22} - \frac{2 x_{0}}{11} - \frac{7}{22}$

So our answer is then

$y = \frac{x^{2}}{22} - \frac{2 x}{11} - \frac{7}{22}$

Report an Error

Example Question #12 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( -8, 9\right ) \\\text{directrix } y = 12$

Possible Answers:

$\left(x - 9\right)^{2} + \left(x + 8\right)^{2}$

$y = - \frac{x^{2}}{24} - \frac{2 x}{3} - \frac{1}{24}$

$\left(x - 2\right)^{2}$

$y = - \frac{x^{2}}{12} - \frac{4 x}{3} - \frac{1}{12}$

$y = - \frac{x^{2}}{6} - \frac{8 x}{3} - \frac{1}{6}$

Correct answer:

$y = - \frac{x^{2}}{6} - \frac{8 x}{3} - \frac{1}{6}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -8 for a 9 for b and 12 for y

$y_{0}^{2} - 24 y_{0} + 144 = x_{0}^{2} + 16 x_{0} + y_{0}^{2} - 18 y_{0} + 145$

Now we can simplify, and solve for $y_{0}$

$y_{0} = - \frac{x_{0}^{2}}{6} - \frac{8 x_{0}}{3} - \frac{1}{6}$

So our answer is then

$y = - \frac{x^{2}}{6} - \frac{8 x}{3} - \frac{1}{6}$

Report an Error

Example Question #13 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( -6, 1\right ) \\\text{directrix } y = -5$

Possible Answers:

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

$\left(x - 2\right)^{2}$

$\left(x - 1\right)^{2} + \left(x + 6\right)^{2}$

$y = \frac{x^{2}}{48} + \frac{x}{4} + \frac{1}{4}$

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{1}{2}$

Correct answer:

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

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -6 for a 1 for b and -5 for y

$y_{0}^{2} + 10 y_{0} + 25 = x_{0}^{2} + 12 x_{0} + y_{0}^{2} - 2 y_{0} + 37$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{12} + x_{0} + 1$

So our answer is then

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

Report an Error

Example Question #14 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( -6, 6\right ) \\\text{directrix } y = -6$

Possible Answers:

$y = \frac{x^{2}}{48} + \frac{x}{4} + \frac{3}{4}$

$y = \frac{x^{2}}{96} + \frac{x}{8} + \frac{3}{8}$

$\left(x - 2\right)^{2}$

$\left(x - 6\right)^{2} + \left(x + 6\right)^{2}$

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{3}{2}$

Correct answer:

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{3}{2}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -6 for a 6 for b and -6 for y

$y_{0}^{2} + 12 y_{0} + 36 = x_{0}^{2} + 12 x_{0} + y_{0}^{2} - 12 y_{0} + 72$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{24} + \frac{x_{0}}{2} + \frac{3}{2}$

So our answer is then

$y = \frac{x^{2}}{24} + \frac{x}{2} + \frac{3}{2}$

Report an Error

Example Question #1 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (9,3) and (-5,3) with a major axis distance of .

Possible Answers:

$10 x^{2} + 10 y^{2} = 1$

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

$\frac{1}{25} \left(x - 2\right)^{2} - \frac{1}{24} \left(y - 3\right)^{2} = 1$

$\frac{1}{10} \left(x - 9\right)^{2} + \frac{1}{10} \left(y - 3\right)^{2} = 1$

$\frac{x^{2}}{10} + \frac{y^{2}}{10} = 1$

Correct answer:

$\frac{1}{25} \left(x - 2\right)^{2} - \frac{1}{24} \left(y - 3\right)^{2} = 1$

Explanation:

The general equation is

$\frac{1}{b^{2}} \left(- h + y\right)^{2} + \frac{1}{a^{2}} \left(x - 9\right)^{2} = 1$

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

{a} 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 = 10

Now solve for

a = 5.0

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

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

$\\x = 2 \\y = 3.0$

So the center of the foci is at $\left ( 2, 3\right )$ .

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 5.0 and with -7

$b^{2} = -24.0$

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

$\frac{1}{25} \left(x - 2\right)^{2} - \frac{1}{24} \left(y - 3\right)^{2} = 1$

Report an Error

Example Question #2 : 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 (-7, 1) with a major axis distance of .

Possible Answers:

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

$\frac{x^{2}}{81} + \frac{1}{32} \left(y - 1\right)^{2} = 1$

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

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

$\frac{1}{18} \left(x - 7\right)^{2} + \frac{1}{18} \left(y - 1\right)^{2} = 1$

Correct answer:

$\frac{x^{2}}{81} + \frac{1}{32} \left(y - 1\right)^{2} = 1$

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 .

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.

$\\x = 0 \\y = 1.0$

So the center of the foci is at $\left ( 0, \quad 1\right )$

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 9.0 and with -7

$b^{2} = 32.0$

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

$\frac{x^{2}}{81} + \frac{1}{32} \left(y - 1\right)^{2} = 1$

Report an Error

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

Find the equation of an ellipse that has foci at (3, 4) and (-5, 4) with a major axis distance of .

Possible Answers:

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

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

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

$\frac{1}{18} \left(x - 3\right)^{2} + \frac{1}{18} \left(y - 4\right)^{2} = 1$

$\frac{1}{81} \left(x + 1\right)^{2} + \frac{1}{65} \left(y - 4\right)^{2} = 1$

Correct answer:

$\frac{1}{81} \left(x + 1\right)^{2} + \frac{1}{65} \left(y - 4\right)^{2} = 1$

Explanation:

The general equation is

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

$\left ( k, \quad h\right )$ 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.

$\\x = -1 \\y = 4.0$

So the center of the foci is at $\left ( -1, \quad 4\right )$

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 = -4

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 -4

$b^{2} = 65.0$

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

$\frac{1}{81} \left(x + 1\right)^{2} + \frac{1}{65} \left(y - 4\right)^{2} = 1$

Report an Error

Example Question #4 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Find the equation of an ellipse that has foci at (-3, 5) and (-1, 5) with a major axis distance of .

Possible Answers:

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

$\frac{1}{64} \left(x + 2\right)^{2} + \frac{1}{63} \left(y - 5\right)^{2} = 1$

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

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

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

Correct answer:

$\frac{1}{64} \left(x + 2\right)^{2} + \frac{1}{63} \left(y - 5\right)^{2} = 1$

Explanation:

The general equation is

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

$\left ( k, \quad h\right )$ 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.

$\\x = -2 \\ y = 5.0$

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

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 = 1

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 1

$b^{2} = 63.0$

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

$\frac{1}{64} \left(x + 2\right)^{2} + \frac{1}{63} \left(y - 5\right)^{2} = 1$

Report an Error

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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 #9 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #10 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #11 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #12 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #13 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #14 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #1 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

Example Question #2 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

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

Example Question #4 : Derive Ellipse And Hyperbola Equations: Ccss.Math.Content.Hsg Gpe.A.3

All Common Core: High School - Geometry Resources

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