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Teddyiswatshecallsme

  • one year ago

NEED HELP! Conics - Circles Write the general conic form equation of each circle. 1. (x + 3)^2 + (y + 6)^2 = 81 2. (x + 6)^2 + (y - 2) = 64 3. (x + 14)^2 + y^2 = 24

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  1. Teddyiswatshecallsme
    • one year ago
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    Please provide a thorough explanation of how you're doing them. I need to understand this stuff.

  2. hwyl
    • one year ago
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    hmm

  3. jdoe0001
    • one year ago
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    \(\bf (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad radius={\color{purple}{ r}} \\ \quad \\ \quad \\ hint:\qquad (x {\color{brown}{ +3}})^2 + (y {\color{blue}{ +6}})^2 = {\color{purple}{ 81}}\)

  4. jdoe0001
    • one year ago
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    hmmm kinda misread.. the genercal conic hmmm

  5. jdoe0001
    • one year ago
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    hint: divide the left side, by the right side

  6. Teddyiswatshecallsme
    • one year ago
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    Huh?

  7. jdoe0001
    • one year ago
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    \(\bf (x + 3)^2 + (y + 6)^2 = 81 \implies \cfrac{(x + 3)^2}{81} + \cfrac{(y + 6)^2}{81} = \cfrac{\cancel{81}}{\cancel{81}} \\ \quad \\ \textit{keeping in mind } \\ \quad \\ \cfrac{(x-{\color{brown}{ h}})^2}{{\color{purple}{ a}}^2}+\cfrac{(y-{\color{blue}{ k}})^2}{{\color{purple}{ b}}^2}=1 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad vertices\ ({\color{brown}{ h}}\pm a, {\color{blue}{ k}}) \)

  8. jdoe0001
    • one year ago
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    also keep in mind that +3 = - (-3)

  9. jim_thompson5910
    • one year ago
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    @jdoe0001 I think they want to expand everything out? For example, (x + 3)^2 = x^2 + 6x + 9

  10. Teddyiswatshecallsme
    • one year ago
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    Now what?

  11. anonymous
    • one year ago
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    Dat sit!! They ask you to get the conic form, just divide both side by the right hand side and you get the conic form of a circle.

  12. Teddyiswatshecallsme
    • one year ago
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    I'm not getting it. o.o

  13. anonymous
    • one year ago
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    To get more details, you can say that that is the "ellipse" with vertices are (\(0,\pm 9\) ) and (\(\pm9, 0)\) Foci (0,0)

  14. jdoe0001
    • one year ago
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    @jim_thompson5910 hmmm methinks is the conic version of it, rather than an expanded one

  15. anonymous
    • one year ago
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    Why not? Like parabola, you have 2 different forms: Standard form and vertex form. Same as circle.

  16. anonymous
    • one year ago
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    @jdoe0001 you are invisible to me again. :)

  17. jdoe0001
    • one year ago
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    hmmm odd

  18. jdoe0001
    • one year ago
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    anyhow @Teddyiswatshecallsme there, make it like an ellipse equation keep in mind that +3 = -(-3) and that +6 = -(-6)

  19. anonymous
    • one year ago
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    Sorry, parabola has 3 forms, polar form also. :)

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