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Math Help - Find the equation of an ellipse

  1. #1
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    Find the equation of an ellipse

    Find the equation of the ellipse with center (-3,-2), a focus at (-3,1), and a vertex at (-3,3).

    What do I do first?
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  2. #2
    Senior Member topher0805's Avatar
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    \frac{(x-h)^{2}}{b^{2}} + \frac{(y-k)^{2}}{a^{2}} = 1

    This is the general form for the equation of the ellipse When the ellipse is vertical. When it is horizontal simply switch the a and the b. The ellipse is centered at the point (h,k), so plug in those values:

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

    Can you solve it from there?
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  3. #3
    Senior Member topher0805's Avatar
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    Hint: a is the distance from the center to the vertex.
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  4. #4
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    What would I be solving it for? What variable?
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  5. #5
    Senior Member topher0805's Avatar
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    I gave you this equation:

    <br />
\frac{(x+3)^{2}}{b^{2}} + \frac{(y+2)^{2}}{a^{2}} = 1<br />

    And I gave you this hint:

    a is the distance from the center to the vertex.

    Now all you have to do is plug in values for a and b.
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  6. #6
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    d = \sqrt{(-3-0)^2 + (3-0)^2}

    d=4.24

    \frac{(x+3)^{2}}{b^{2}} + \frac{(y+2)^{2}}{4.24^{2}} = 1<br />

    Then Solve for b^2 ?
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  7. #7
    Senior Member topher0805's Avatar
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    Hmm, how did you get 4.24?

    Find the correct value for a (it should be 5) and then solve for b. You will need to know a point on the ellipse, but thankfully you already have one!


    I haven't done these in a long time so forgive me if I am a bit rusty.
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  8. #8
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    \frac{(-3+3)^{2}}{b^{2}} + \frac{(1+2)^{2}}{5^{2}} = 1

    This the right equation so far?

    Edit: I'm lost because that would make it \frac{0}{b^2} which is 0. What did I do wrong?
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  9. #9
    Senior Member topher0805's Avatar
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    Sorry, I just realized that you don't have a point on the ellipse. Instead you will have to use trig to find b.

    You have a focus (-3,1). The definition of a focus is that the distance from the two focus' to any point on the ellipse is constant.

    Eg. If you have point A and point B on the ellipse, the distance from f1 to A to f2 will be the same as f1 to B to f2. Am I explaining this clearly?

    Since your one focus is (-3,1), your other will be (-3, -5). The distance from one focus to the vertex given, to the second focus is 10. Since the two focuses are equidistant from each other, the point b is therefore 5 away from both focuses. Use Pythagoras to solve for b in this manner:

    5^2 = b^2 + 4^2
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  10. #10
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    Hrmm...That is some complex stuff. I hate graphs so much.
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  11. #11
    Senior Member topher0805's Avatar
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    Yeah, I can draw some graphs and show a better way to do it and post it here if you like but right now I'm trying to get my homework done so it won't be up till morning.
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  12. #12
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    \frac{(x+3)^{2}}{9} + \frac{(y+2)^{2}}{25} = 1

    Final Equation. Right?
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  13. #13
    Senior Member topher0805's Avatar
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    Yes that is correct.

    You'll be a pro in no time.
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