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anonymous

  • one year ago

I'm new to this website and how it works, but I would appreciate it if somebody helped me! Find the standard form of the equation of the parabola with a focus at (7, 0) and a directrix at x = -7.

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  1. PhantomCrow
    • one year ago
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    What did you notice when you graphed these two points on a plane?

  2. anonymous
    • one year ago
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    @Phantomcrow It opens to the right?

  3. PhantomCrow
    • one year ago
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    Yes! And you know what the standard equation of a parabola that opens to the right is?

  4. anonymous
    • one year ago
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    @Phantomcrow Can you help me out with that one

  5. anonymous
    • one year ago
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    It is 3, you're welcome.

  6. PhantomCrow
    • one year ago
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    Sure.|dw:1444091089825:dw|

  7. PhantomCrow
    • one year ago
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    Since the vertex is at the origin, there is no need to add any constants for shifts. The graph is simply \[y^2=x\]

  8. anonymous
    • one year ago
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    @Phantomcrow By the way the Answer choices are y= 1/28x^2 x= 1/28y^2 -28y= x^2 y^2=14x What confuses me is how these tie to the standard formula

  9. PhantomCrow
    • one year ago
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    D. Despite it having an 'x' term, its vertex is still at the origin.

  10. PhantomCrow
    • one year ago
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    Standard form is one of the ways to write the graph of a parabola. It is commonly seen as:\[ax^2+bx+c\]

  11. anonymous
    • one year ago
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    @Phantomcrow Wow thanks, would it be cool if you helped me with one more?

  12. PhantomCrow
    • one year ago
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    Sure

  13. anonymous
    • one year ago
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    @Phantomcrow Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8, i'm guessing this is similar or even exact to the last one, so it opens to the left

  14. PhantomCrow
    • one year ago
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    Actually, it does not open to the left this time. You are given y values for focus and directrix, not x values.

  15. anonymous
    • one year ago
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    @Phantomcrow So y values open to the right and x values open to the left?

  16. PhantomCrow
    • one year ago
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    Y values lie across a vertical line, you your parabola will be opening upwards. You can easier see this if you look at the (0.8) and y= -8 on a graph. See that the they are equidistant from a point.

  17. anonymous
    • one year ago
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    @PhantomCrow you graph (0,8) correct?

  18. PhantomCrow
    • one year ago
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    Yes. Graph the point (0,8) and y= -8 and you should be able to visually see the the point that is equidistant from them. There is a formula for finding distance but I believe simply looking at the graph for this problem will provide you with the answer.

  19. anonymous
    • one year ago
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    @PhantomCrow The choices are y= 1/32x^2 y^2= 8x y^2=32x y=1/8x^2 Would it be B?

  20. PhantomCrow
    • one year ago
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    Oh, I see.

  21. PhantomCrow
    • one year ago
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    Well, you have two possible answers. A and D. B and C are ruled out because that would be a parabola who's focus is (x,0) and directrix is x=something. We have a focus like (0,y) and y=something in this case so its not y^2=x

  22. PhantomCrow
    • one year ago
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    Since you need to figure out which is correct between A and D, your best method is to plug in what you know into vertex form (\[(x-h)^2=4p(y-k)\] where (h, k) is the vertex and p is the distance from either the directrix or the focus to the vertex.

  23. PhantomCrow
    • one year ago
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    We know that the vertex lies at (0,0) because that is the point equidistant from (0,8) and y=-8. P is not difficult to find. P is just 8 because the distance from either the directrix or the focus to the vertex is 8 units (if you need the formula, just ask, but its not necessary).

  24. PhantomCrow
    • one year ago
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    Plug that into the vertex form:\[(x-h)^2=4p(y-k)\] \[(h,k)=(0,0)\] so \[h=0, k=0\] \[(x-0)^2=4(8)(y-k)\] \[x^2=32y\] \[\frac{ x^2 }{ 32 }=y\]

  25. PhantomCrow
    • one year ago
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    We want it in terms of y because that's how standard form is written. So we solved for y.

  26. anonymous
    • one year ago
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    @PhantomCrow Which will result in y=1/32x^2

  27. PhantomCrow
    • one year ago
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    Yep.

  28. anonymous
    • one year ago
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    @PhantomCrow Thank you so much for your help have a blessed day!

  29. PhantomCrow
    • one year ago
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    No problem, you as well.

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