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Flashcards: Derive Ellipse and Hyperbola Equations: CCSS.Math.Content.HSG-GPE.A.3

Find the equation of an ellipse that has foci at (9,3) $\displaystyle (9,3)$ and (-5,3) $\displaystyle (-5,3)$ with a major axis distance of $\displaystyle 10$.

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

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

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

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

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

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Flashcards: Derive Ellipse and Hyperbola Equations: CCSS.Math.Content.HSG-GPE.A.3

Find the equation of an ellipse that has foci at (9,3) $\displaystyle (9,3)$ and (-5,3) $\displaystyle (-5,3)$ with a major axis distance of $\displaystyle 10$.

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

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

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

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

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

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Flashcards: Derive Ellipse and Hyperbola Equations: CCSS.Math.Content.HSG-GPE.A.3

Find the equation of an ellipse that has foci at (9,3) $\displaystyle (9,3)$ and (-5,3) $\displaystyle (-5,3)$ with a major axis distance of $\displaystyle 10$.

Your answer:

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

Your answer:

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

Your answer:

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

Your answer:

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

Correct answer:

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

Explanation:

The general equation is

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

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

{a} $\displaystyle {a}$ is the distance to the foci on the x-axis from the center, and $\displaystyle b$ is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is $\displaystyle a$.

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

$\displaystyle 2 a = 10$

Now solve for $\displaystyle a$

a = 5.0 $\displaystyle a = 5.0$

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

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

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

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

Now we need to figure out what the distance from the center to the foci is $\displaystyle c$.

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

c = -7 $\displaystyle c = -7$

Now the last part is to find $\displaystyle b$.

We can find it by using the following equation.

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

We simply substitute $\displaystyle a$ with 5.0 and $\displaystyle c$ with -7

$b^{2} = -24.0$ $\displaystyle b^{2} = -24.0$

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

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

Flashcards: Derive Ellipse and Hyperbola Equations: CCSS.Math.Content.HSG-GPE.A.3

Find the equation of an ellipse that has foci at (7, 1) $\displaystyle (7, 1)$ and (-7, 1) $\displaystyle (-7, 1)$ with a major axis distance of $\displaystyle 18$.

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

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

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

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

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

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Flashcards: Derive Ellipse and Hyperbola Equations: CCSS.Math.Content.HSG-GPE.A.3

Find the equation of an ellipse that has foci at (7, 1) $\displaystyle (7, 1)$ and (-7, 1) $\displaystyle (-7, 1)$ with a major axis distance of $\displaystyle 18$.

Your answer:

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

Your answer:

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

Your answer:

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

Your answer:

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

Correct answer:

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

Explanation:

The general equation is

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

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

$\displaystyle a$ is the distance to the foci on the x-axis from the center, and $\displaystyle b$ is the distance to the foci from the y-axis.

The first step is to figure out what the focal radii is $\displaystyle a$.

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

$\displaystyle 2 a = 18$

Now solve for $\displaystyle a$.

a = 9.0 $\displaystyle a = 9.0$

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

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

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

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

Now we need to figure out what the distance from the center to the foci is $\displaystyle c$.

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

c = -7 $\displaystyle c = -7$

Now the last part is to find $\displaystyle b$.

We can find it by using the following equation.

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

We simply substitute $\displaystyle a$ with 9.0 and $\displaystyle c$ with -7

$b^{2} = 32.0$ $\displaystyle b^{2} = 32.0$

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

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

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Flashcards: Derive Ellipse and Hyperbola Equations: CCSS.Math.Content.HSG-GPE.A.3

All Common Core: High School - Geometry Resources

Flashcards: Derive Ellipse and Hyperbola Equations: CCSS.Math.Content.HSG-GPE.A.3

Flashcards: Derive Ellipse and Hyperbola Equations: CCSS.Math.Content.HSG-GPE.A.3

Flashcards: Derive Ellipse and Hyperbola Equations: CCSS.Math.Content.HSG-GPE.A.3

Flashcards: Derive Ellipse and Hyperbola Equations: CCSS.Math.Content.HSG-GPE.A.3

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