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Math Help - Complete the square.. and.

  1. #1
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    Complete the square.. and.

    Okay I have to "little" problems to do like my teacher said .


    1.) Complete the square to write the equation in standard form and sketch the graph.

    x^2+4x+3y^2-5=0


    2.) Determine an equation for an ellipse with center (5,2), semi major axis = 3, and one focus at (7,-2).
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  2. #2
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    Quote Originally Posted by Bradley55 View Post

    1.) Complete the square to write the equation in standard form and sketch the graph.

    x^2+4x+3y^2-5=0

    ...
    x^2+4x+3y^2-5=0~\implies~x^2+4x+{\color{blue} 4} +3y^2=5+{\color{blue}4} Now divide both sides by 9:

    \frac{(x+2)^2}{9} + \frac{y^2}{3}=1

    That's the equation of an ellipse with center at C(-2, 0) and the semi-axes a = 3 and b = \sqrt{3}
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  3. #3
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    Quote Originally Posted by Bradley55 View Post
    ...


    2.) Determine an equation for an ellipse with center (5,2), semi major axis = 3, and one focus at (7,-2).
    I assume that there is a typo:

    You are supposed to know that the distance between the center and one focus of the ellipse is called the linear excentricity e. It is calculated by:

    e^2+b^2=a^2

    With your problem e^2=20 that means it is greater than a so a can't be the semi major axis.

    The points C(5, 2) and F(7, -2) indicate an ellipse which has been rotated by 63.4 clockwise.

    Before I do the calculations please check the wording of your problem.
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  4. #4
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    Ooops.

    The type was

    (5,-2)

    question again is.


    Determine an equation for an ellipse with center (5,-2), semimajor axis = 3, and one focus at (7,2).
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  5. #5
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    Quote Originally Posted by Bradley55 View Post
    ...

    Determine an equation for an ellipse with center (5,-2), semimajor axis = 3, and one focus at (7,-2).
    The distance between the center and the focus is 2, that means e = 2.

    Therefore:

    2^2+b^2=3^2~\implies~b=\sqrt{5}

    Plug in all the values you know into the general equation of an ellipse:

    \frac{(x-5)^2}{3^2}+\frac{(y+2)^2}{(\sqrt{5})^2}=1 ~\implies~ \frac{(x-5)^2}{9}+\frac{(y+2)^2}{5}=1
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