anonymous
  • anonymous
HELP PLEASE!!!! find the equation of ellipse. center @origin. 1. Passing through (-2, 9/4), vertex at (4,0) 2.passing through (2, - square root 2) latus rectum square root three.. 4. Passing through (2, square root 30 / 3.), distance between directrices 24 square root 7 / 7. 6. Passing through (0, square root 5), eccentricity equal to 4 square root 21 / 21. 8. Distance between directrices equal to 9 square root 2 / two, latus rectum euqal to 2/3 9. Find the eccentricity and equation of the ellipse if the distance between its directrices is 9 square root 14 / 14 times its minor axis, length of latus rectum 4/3 10. Find the eccentricity and equation of the ellipse if the latus rectum is five-sixteenths that of the major axis, the minor axis equal to 2 square root 5. Please answer please..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
|dw:1378893882422:dw|
anonymous
  • anonymous
Equation of an ellipse:\[ \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2 = 1 \]

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anonymous
  • anonymous
yes yes..
anonymous
  • anonymous
We already know \(a=4\).
anonymous
  • anonymous
Prodigious!
anonymous
  • anonymous
how about b??
anonymous
  • anonymous
Plug in \((-2,9/4)\):\[ \left(\frac{-2}{4}\right)^2+\left(\frac{9/4}{b}\right)^2 = 1 \] Solve for \(b\).
anonymous
  • anonymous
yes! thank you so much. how about the second condition?
anonymous
  • anonymous
\[ a^2+b^2=c^2 = (\sqrt{3})^2=3 \]
anonymous
  • anonymous
And so: \(b^2=3-a^2\)
anonymous
  • anonymous
\[ \frac{2^2}{a^2}+\frac{(-\sqrt{2})^2}{3-a^2}=1 \]Solve for \(a\). Plug it back in to get \(b\).
anonymous
  • anonymous
i forgot.. the answer in the first condition must be 27x^2 + 64y^2 = 432.. im so sorry..
anonymous
  • anonymous
6. Passing through (0, square root 5), eccentricity equal to 4 square root 21 / 21 you can find answer please tell me

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