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ariesvang

  • 2 years ago

find the vertices for the ellipse (x+1)^2 + 4(y+3)^2 = 9

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  1. Mashy
    • 2 years ago
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    standard form of an ellipse with vertices at h and k is given by \[(x-h)^{2}/a ^{2} + (y-k)^{2}/b ^{2}= 1\]

  2. Mashy
    • 2 years ago
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    so compare and get the answer !

  3. ariesvang
    • 2 years ago
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    so the vertices are h and k right??

  4. Mashy
    • 2 years ago
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    Yes in that given general equation its, h, k.. so in your equation what is h and k?

  5. ariesvang
    • 2 years ago
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    it would be 1 and 3

  6. Mashy
    • 2 years ago
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    No.. check properly.. if it was 1 and 3.. the equation to the ellipse would be (x-1)^2 + 4(y-3)^2 = 9

  7. Mashy
    • 2 years ago
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    so.. what is it??

  8. ariesvang
    • 2 years ago
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    it would be -1 and -3 ??

  9. Mashy
    • 2 years ago
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    Bingo!

  10. ariesvang
    • 2 years ago
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    so thats the vertices??and thats it??

  11. Mashy
    • 2 years ago
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    Yup that's it.. kind of a puny question eh?? Go ask your teachers to give you better problems to solve :P

  12. phi
    • 2 years ago
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    You should check your definitions (google works) See http://www.mathwords.com/v/vertices_of_an_ellipse.htm for what the vertex is. You found the center of the ellipse

  13. phi
    • 2 years ago
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    Your equation is (x+1)^2 + 4(y+3)^2 = 9 in standard form this is \[ \frac{(x+1)^2}{3^2}+\frac{(y+3)^2}{(\frac{3}{2})^2}=1 \] The semi-major axis is the longest axis. Here it is 3 (denominator of x term) the center is (-1,-3) and when y= -3 you are on the semi-major axis. x= -1+3= 2 or x= -1-3= -4 and the vertices are (-4,-3) and (2,-3)

  14. eliassaab
    • 2 years ago
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    Here is a graph of it

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  15. ariesvang
    • 2 years ago
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    thank you both :)

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