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alyssababy7

Find the standard form of the equation of the ellipse satisfying the given conditions. Foci: (0, -2), (0, 2); y-intercepts: -5 and 5

  • one year ago
  • one year ago

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  1. satellite73
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    since the foci are at \((0,-2)\) and \((0,2)\) the center is at the origin, so at least we know it looks like \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\]

    • one year ago
  2. satellite73
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    since the vertices are at \((0,-5)\) and \((0,5)\) we know \(b=5\) and so \(b^2=25\)

    • one year ago
  3. alyssababy7
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    Thank you

    • one year ago
  4. satellite73
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    we still need \(a^2\) but since \(2^2=b^2-a^2\) we can solve for it

    • one year ago
  5. satellite73
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    \(a^2=25-4=21\)

    • one year ago
  6. satellite73
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    let me make sure this is right before i tell you something silly

    • one year ago
  7. satellite73
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    yes, i think this works, i.e. the answer would be \[\frac{x^2}{21}+\frac{y^2}{25}=1\]

    • one year ago
  8. satellite73
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    lets check it http://www.wolframalpha.com/input/?i=ellipse+x^2%2F21%2By^2%2F25%3D1

    • one year ago
  9. alyssababy7
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    i had a multiple choice answer, and when you told me 25 was b^2, it was the only option. you are right, it is x^2/21 + y^2/25 = 1

    • one year ago
  10. satellite73
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    oh, hiding that information... well it works in any case

    • one year ago
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