Practice Problems

Answers

  1. The strategy

    In order to find the equation of the ellipse, we perform the following steps.

    1. Find the focal length, f, and the major radius,‍ p, using the given information about the center, the foci, and the vertices.

    2. Find the minor radius, q, using the equation f^2=p^2-q^2.

    3. Match the major and minor radii to the vertical and horizontal radii by determining the axis along which the foci lie.

    4. Substitute all the values we found in the standard equation of an ellipse.

    Finding the focal length and major radius

    The focal length is 5 units, since the foci are at ‍(\pm 5,0).

    The major radius is \sqrt{41} units, since the vertices are at ‍(\pm \sqrt{41},0).

    Finding the minor radius

    We've found that the focal length, f, is ‍5 units, and the major radius, ‍p, is ‍\sqrt{41} units. Let's substitute these values into the equation to find the minor radius, q.

    \begin{aligned}f^2&=p^2-q^2\\\\5^2&=(\sqrt{41})^2-q^2\\\\41-25&=q^2\\\\4&=q\end{aligned}
     
    Therefore, the minor radius is 4 units.

    Matching the major and minor radii with the vertical and horizontal radii

    Since the foci are located on the x-axis, the major radius is the horizontal radius.

    In consequence, the minor radius is the vertical radius.

    This means that the vertical radius is 4 units and the horizontal radius is \sqrt{41} units.

    Writing the equation

    Our ellipse is centered at (0, 0), has a horizontal radius of ‍{\sqrt{41}} units and a vertical radius of ‍4 units. So it can be represented by the equation below.

    \begin{aligned}\dfrac{(x - 0)^2}{( {\sqrt{41}})^2} + \dfrac{(y - 0)^2}{{4}^2} &= 1\\\\\\\dfrac{x^2}{{41}} + \dfrac{y^2}{{16}} &= 1 &(\text{Simplify terms})\end{aligned}

    Summary

    The ellipse in question can be represented by the following equation.

    \dfrac{x^2}{41} + \dfrac{y^2}{ 16} = 1

  2. The strategy

    In order to find the equation of the ellipse, we perform the following steps.

    1. Find the focal length, f, and the major radius,‍ q, using the given information about the center, the foci, and the vertices.

    2. Find the minor radius, p, using the equation f^2=p^2-q^2.

    3. Match the major and minor radii to the vertical and horizontal radii by determining the axis along which the foci lie.

    4. Substitute all the values we found in the standard equation of an ellipse.

    Finding the focal length and major radius

    The focal length is \sqrt{13} units, since the foci are at ‍(\pm \sqrt{13},0).

    The major radius is 11 units, since the vertices are at ‍(0,\pm 11).

    Finding the minor radius

    We've found that the focal length, f, is ‍\sqrt{13} units, and the major radius, ‍q, is ‍11 units. Let's substitute these values into the equation to find the minor radius, p.

    \begin{aligned}f^2&=p^2-q^2\\\\(\sqrt{13})^2&=p^2-11^2\\\\121+13&=p^2\\\\\sqrt{134}&=p\end{aligned}
     
    Therefore, the minor radius is \sqrt{134} units.

    Matching the major and minor radii with the vertical and horizontal radii

    Since the foci are located on the x-axis, the major radius is the horizontal radius.

    In consequence, the minor radius is the vertical radius.

    This means that the vertical radius is 11 units and the horizontal radius is \sqrt{134} units.

    Writing the equation

    Our ellipse is centered at (0, 0), has a horizontal radius of ‍\sqrt{134} units and a vertical radius of ‍11 units. So it can be represented by the equation below.

    \begin{aligned}\dfrac{(x - 0)^2}{( {\sqrt{134}})^2} + \dfrac{(y - 0)^2}{{11}^2} &= 1\\\\\\\dfrac{x^2}{{134}} + \dfrac{y^2}{{121}} &= 1 &(\text{Simplify terms})\end{aligned}

    Summary

    The ellipse in question can be represented by the following equation.

    \dfrac{x^2}{134} + \dfrac{y^2}{ 121} = 1

  3. The strategy

    In order to find the equation of the ellipse, we perform the following steps.

    1. Find the focal length, f, and the major radius,‍ p, using the given information about the center, the foci, and the vertices.

    2. Find the minor radius, q, using the equation f^2=p^2-q^2.

    3. Match the major and minor radii to the vertical and horizontal radii by determining the axis along which the foci lie.

    4. Substitute all the values we found in the standard equation of an ellipse.

    Finding the focal length and major radius

    The focal length is 8 units, since the foci are at ‍(\pm 8,0).

    The major radius is 17 units, since the vertices are at ‍(\pm 17,0).

    Finding the minor radius

    We've found that the focal length, f, is ‍8 units, and the major radius, ‍p, is ‍17 units. Let's substitute these values into the equation to find the minor radius, q.

    \begin{aligned}f^2&=p^2-q^2\\\\8^2&=17^2-q^2\\\\289-64&=q^2\\\\15&=q\end{aligned}
     
    Therefore, the minor radius is 15 units.

    Matching the major and minor radii with the vertical and horizontal radii

    Since the foci are located on the x-axis, the major radius is the horizontal radius.

    In consequence, the minor radius is the vertical radius.

    This means that the vertical radius is 15 units and the horizontal radius is 17 units.

    Writing the equation

    Our ellipse is centered at (0, 0), has a horizontal radius of ‍17 units and a vertical radius of ‍15 units. So it can be represented by the equation below.

    \begin{aligned}\dfrac{(x - 0)^2}{{17}^2} + \dfrac{(y - 0)^2}{{15}^2} &= 1\\\\\\\dfrac{x^2}{{289}} + \dfrac{y^2}{{225}} &= 1 &(\text{Simplify terms})\end{aligned}

    Summary

    The ellipse in question can be represented by the following equation.

    \dfrac{x^2}{289} + \dfrac{y^2}{ 225} = 1

  4. The strategy

    In order to find the equation of the ellipse, we perform the following steps.

    1. Find the focal length, f, and the major radius,‍ p, using the given information about the center, the foci, and the vertices.

    2. Find the minor radius, q, using the equation f^2=p^2-q^2.

    3. Match the major and minor radii to the vertical and horizontal radii by determining the axis along which the foci lie.

    4. Substitute all the values we found in the standard equation of an ellipse.

    Finding the focal length and major radius

    The focal length is \sqrt{8} units, since the foci are at ‍ (\pm \sqrt{8},0) .

    The major radius is \sqrt{10} units, since the vertices are at ‍(0,\pm \sqrt{10}).

    Finding the minor radius

    We've found that the focal length, f, is ‍\sqrt{8} units, and the major radius, ‍p, is ‍\sqrt{10} units. Let's substitute these values into the equation to find the minor radius, q.

    \begin{aligned}f^2&=p^2-q^2\\\\(\sqrt{8})^2&=p^2-(\sqrt{10})^2\\\\8+10&=p^2\\\\\sqrt{18}&=p\end{aligned}
     
    Therefore, the minor radius is \sqrt{18} units.

    Matching the major and minor radii with the vertical and horizontal radii

    Since the foci are located on the x-axis, the major radius is the horizontal radius.

    In consequence, the minor radius is the vertical radius.

    This means that the vertical radius is \sqrt{10} units and the horizontal radius is \sqrt{18} units.

    Writing the equation

    Our ellipse is centered at (0, 0), has a horizontal radius of ‍\sqrt{18} units and a vertical radius of ‍\sqrt{10} units. So it can be represented by the equation below.

    \begin{aligned}\dfrac{(x - 0)^2}{( {\sqrt{18}})^2} + \dfrac{(y - 0)^2}{({\sqrt{10}})^2} &= 1\\\\\\\dfrac{x^2}{{18}} + \dfrac{y^2}{{10}} &= 1 &(\text{Simplify terms})\end{aligned}

    Summary

    The ellipse in question can be represented by the following equation.

    \dfrac{x^2}{18} + \dfrac{y^2}{ 10} = 1
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