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anonymous

  • one year ago

Derive the equation of the parabola with a focus at (4, −7) and a directrix of y = −15. Put the equation in standard form.

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  1. campbell_st
    • one year ago
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    ok... have you plotted the information on the number plane so you know the concavity of the parabola..?

  2. anonymous
    • one year ago
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    yeah

  3. anonymous
    • one year ago
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    So, you can still use the combined distance equation for this problem. But you just have to make your solution ax^2+bx+c=0.

  4. anonymous
    • one year ago
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    ax+by=c.

  5. campbell_st
    • one year ago
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    so is the directrix above or below the focus...? and then is the parabola concave up or down..?

  6. campbell_st
    • one year ago
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    |dw:1438892282167:dw| so you plot looks like this

  7. anonymous
    • one year ago
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    below right?

  8. campbell_st
    • one year ago
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    |dw:1438892378795:dw| the distance is labelled 2a and is double the focal length.

  9. anonymous
    • one year ago
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    8

  10. campbell_st
    • one year ago
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    the dotted line through x = 4 is the line of symmetry.... and the vertex is on this line midway between the focus and directrix

  11. campbell_st
    • one year ago
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    great so if the twice the focal length is 8 then the focal length is a = 4 so the vertex is 4 units below the focus on the line x = 4 so where do you think the vertex is..?

  12. campbell_st
    • one year ago
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    |dw:1438892601545:dw|

  13. anonymous
    • one year ago
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    at y=-11 ??

  14. anonymous
    • one year ago
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    or no

  15. campbell_st
    • one year ago
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    great... so the standard form I use now is \[(x - h)^2 = 4a(y - k)\] (h, k) is the vertex and a is the focal length so the equation is \[(x -4)^2 = 4 \times 4(y + 11)\] now you just need to simplify this equation

  16. anonymous
    • one year ago
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    f(x)=1/16x^-8x+11 ?

  17. campbell_st
    • one year ago
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    so if you want it in vertex form its \[\frac{1}{16}(x -4)^2 = y + 11\] then subtract 11 from both sides

  18. campbell_st
    • one year ago
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    well not quite \[\frac{1}{16}(x^2 - 8x + 16)- 11 = y\]

  19. campbell_st
    • one year ago
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    which would become \[y = \frac{x^2}{16} - \frac{x}{2} - 10 \]

  20. anonymous
    • one year ago
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    ohh ok, thats the same as 1/16x^2-1/2x-10 right?

  21. campbell_st
    • one year ago
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    it is the same

  22. anonymous
    • one year ago
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    ok thank you so much :)

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