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calyne Group Title

If c > (1/2), how many lines through the point (0,c) are normal lines to the parabola y = x^2? What if c < or = (1/2)? Please explain the steps...

  • 2 years ago
  • 2 years ago

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  1. dpaInc Group Title
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    is this a calculus question?

    • 2 years ago
  2. dpaInc Group Title
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    if c=1/2 then there is only one line... that is the y-axis. if c<1/2, there are no lines because the parabola will always be positive. Now if c>1/2, there are always two lines.

    • 2 years ago
  3. calyne Group Title
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    how do you figure

    • 2 years ago
  4. dpaInc Group Title
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    take the derivative of y=x^2 to get y'=2x. This gives the slope of the tangent line for what ever x coordinate you look at that's on the parabola. Since you want the line that's normal to the parabola, take the negative reciprocal of the derivative to get you the slope of the line perpendicular to the tangent line. Since you know the line goes through (0, c) you can use that as the point to get the equation of the line perpendicular to tangent line using point-slope form of a line. you should end up with this equation: y = c - 1/2. this is how I came to the conclusions in my previous post.

    • 2 years ago
  5. calyne Group Title
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    so the slope of the normal line is -1/(2x)??

    • 2 years ago
  6. dpaInc Group Title
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    yep

    • 2 years ago
  7. calyne Group Title
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    i thought the slope had to be a number

    • 2 years ago
  8. calyne Group Title
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    like so that's equal to a slope of -1/2?

    • 2 years ago
  9. dpaInc Group Title
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    yes but the slope on a parabola depends on what point you look at. so the slope is given as an expression.

    • 2 years ago
  10. calyne Group Title
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    ohh for real ok

    • 2 years ago
  11. dpaInc Group Title
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    did you get to this point of the problem: y = c- 1/2?

    • 2 years ago
  12. calyne Group Title
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    i don't know i mean y-c = -(1/2x)(x-0)? or -(1/2)(x-0) ???

    • 2 years ago
  13. calyne Group Title
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    ...the second right

    • 2 years ago
  14. dpaInc Group Title
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    yes... you got it now just simplify the right side to -1/2*x

    • 2 years ago
  15. calyne Group Title
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    right obviously so ok then yeah y = c-1/2*x

    • 2 years ago
  16. dpaInc Group Title
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    sorry, i mean -1/2

    • 2 years ago
  17. dpaInc Group Title
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    no x's

    • 2 years ago
  18. calyne Group Title
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    wait what

    • 2 years ago
  19. dpaInc Group Title
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    simplifying the right side, the x's cancell out right? you're left with only -1/2

    • 2 years ago
  20. calyne Group Title
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    where'd the x go

    • 2 years ago
  21. calyne Group Title
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    oh ok so it IS -1/2*x(x-0) right to begin with

    • 2 years ago
  22. dpaInc Group Title
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    just to be more clear, the right side is (-1/(2x))*(x-0)

    • 2 years ago
  23. calyne Group Title
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    right right that's what i meant ok

    • 2 years ago
  24. calyne Group Title
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    thanks

    • 2 years ago
  25. dpaInc Group Title
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    np

    • 2 years ago
  26. calyne Group Title
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    so if c = 1/2, y = 0, and... how does that mean that there is one line and that it's the y-axis?

    • 2 years ago
  27. dpaInc Group Title
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    when y=0, this makes x=0. so the line normal to the parabola at (0,0) is the vertical line, x=0, and this is the only one when c=1/2

    • 2 years ago
  28. calyne Group Title
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    oh.. k. so if c < 1/2, it's negative, and x^2 is always positive on y... how would i check that otherwise though if i didn't know the graph of the function?

    • 2 years ago
  29. dpaInc Group Title
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    from your equation, y=x^2, notice that x^2 is always positive... so y must always be positive.

    • 2 years ago
  30. calyne Group Title
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    but uh i'm sorry i know i must seem stupid as hell but how does that (c = 1/2, y = 0, x = 0, etc) mean that x is always 0 on the line? that's not just a point? geez....

    • 2 years ago
  31. calyne Group Title
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    that i know i was asking like if i didn't know the graph of the function like i know the graph of x^2... but obviously it's no big deal, same deal, it's gotta fit in the domain/range..... but uh there i'm a little fuzzy anyway what the hell is c in the normal line

    • 2 years ago
  32. dpaInc Group Title
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    it does not mean that x is always zero, all I did was to solve y=x^2 when y=0. Now, if c>1/2, notice that your y is positive so that yields two x values, one positive and the other negative.

    • 2 years ago
  33. calyne Group Title
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    alright but so what is that you're finding? how is the resulting x,y a LINE and not a POINT? you know what i mean? that's how i'm confused

    • 2 years ago
  34. calyne Group Title
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    oh they're each other points on the normal line that goes through the point (0,c) so that makes a line

    • 2 years ago
  35. calyne Group Title
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    one for each result i got it sorry

    • 2 years ago
  36. dpaInc Group Title
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    remember the question was "how many lines..." ? the resulting points gave you the other point(s) needed to connect with (0, c) and be normal to the parabola...

    • 2 years ago
  37. calyne Group Title
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    gotcha

    • 2 years ago
  38. dpaInc Group Title
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    nice problem... thank you

    • 2 years ago
  39. calyne Group Title
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    aw no thank you

    • 2 years ago
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