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anonymous

  • one year ago

Can someone help me with my conic sections assignment? I'm really confused.

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  1. anonymous
    • one year ago
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    I have a few that I really don't know how to do, but here's one: What is the equation of the following graph?

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  2. phi
    • one year ago
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    First, what shape is that? is it a circle, ellipse, parabola or hyperbola?

  3. anonymous
    • one year ago
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    ellipse

  4. phi
    • one year ago
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    next, look in your notes for the equation of an ellipse. can you write down the "general form" ?

  5. anonymous
    • one year ago
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    \[\frac{ x^{2} }{ a ^{2}} + \frac{ y ^{2}}{ b ^{2} }\]

  6. phi
    • one year ago
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    an equation has an = sign in it. it should be \[ \frac{ x^{2} }{ a ^{2}} + \frac{ y ^{2}}{ b ^{2} }=1 \] but that is for an ellipse whose center is at (0,0) your ellipse is not centered at (0,0). The better equation is \[ \frac{ (x-h)^{2} }{ a ^{2}} + \frac{ (y-k) ^{2}}{ b ^{2} } =1\] where (h,k) is the center of the ellipse. Can you figure out what point is in the exact center of your ellipse ?

  7. anonymous
    • one year ago
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    (0,-2)

  8. phi
    • one year ago
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    that means you know (h,k) h=0 and k= -2 put those into the equation

  9. anonymous
    • one year ago
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    \[\frac{(x-0)^{2} }{ a ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}}\]

  10. phi
    • one year ago
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    = 1 don't leave that out also, x-0 simplifies to x so \[ \frac{x^{2} }{ a ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}} =1 \]

  11. phi
    • one year ago
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    now for the "a" value: that is the distance in the x-direction (sideways) from the center to the side. what is the value of "a" ?

  12. anonymous
    • one year ago
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    1?

  13. phi
    • one year ago
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    yes, a is 1 and a^2 which means a*a is 1*1 = 1 so \[ \frac{x^{2} }{ a ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}} =1 \\ \frac{x^{2} }{ 1 ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}} =1 \] now find b, which the distance from the center to the top (i.e. in the up/down direction , or y-axis direction)

  14. anonymous
    • one year ago
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    3?

  15. phi
    • one year ago
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    yes

  16. phi
    • one year ago
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    \[ \frac{x^{2} }{ 1 ^{2} }+ \frac{ (y+2)^{2} }{ 3^{2}} =1 \] we could simplify that a little, and write \[ x^2 + \frac{ (y+2)^{2} }{ 9} =1 \] but either way is ok

  17. anonymous
    • one year ago
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    So is that the final answer?

  18. phi
    • one year ago
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    yes. You can check some points to make sure it's correct. for example , when x is 0 you get \[ 0^2 + \frac{ (y+2)^{2} }{ 9} =1 \\ \frac{ (y+2)^{2} }{ 9} =1 \\ (y+2)^2= 9 \\ y+2= \pm 3 \] and y= +3-2= 1 or y= -3-2= -5 so points (0,1) and (0, -5) should be on the ellipse.

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