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Find the equation of a tangent line to y = x^2?

  1. Jul 21, 2017 #1
    1. The problem statement, all variables and given/known data
    the line goes through (0, 3/2) and is orthogonal to a tangent line to the part of parabola y = x^2, x > 0

    2. Relevant equations


    3. The attempt at a solution
    I have problems regarding finding the equation of tangent line to the part of parabola
    because the question not specifically mention at which point

    we know that the tangent line (y2) tangent to the part of parabola
    so, the slope of y2 = the slope of parabola at (x,y)

    parabola y = x^2
    so the slope is 2x
    and the slope changes according to x

    which x should I used for x > 0?
    if I use x = 1, y = 1^2= 1, slope = 2
    and the y2 would be
    y - 1 = 2(x - 1)
    y = 2x - 1

    but if I used x = 3, y = 9, slope = 6
    and the y2 would be
    y - 9 = 6(x-3)
    y = 6x - 9

    and y2 is different for each x...

    I need to find the slope of y2, so I can find the line equation orthogonal to y2 passing through (0, 3/2) ...
     
  2. jcsd
  3. Jul 21, 2017 #2

    scottdave

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    Do you know the relationship between the slopes of perpendicular (orthogonal) lines?
    This article will help. http://www.purplemath.com/modules/slope3.htm

    If you know the relationship between slopes of orthogonal lines, then you can get an expression for the slope of a line orthogonal to the parabola. They give you what point to use.
     
  4. Jul 21, 2017 #3
    for y1 orthogonal to y2
    means
    slope of y1 * slope of y2 = -1

    how to find the equation of a line orthogonal to the parabola?
    y - y1 = m(x - x1)
    what's the slope? and which point?

    slope of parabola = 2x
     
  5. Jul 21, 2017 #4

    CWatters

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    A tangent to a curve has the same slope as the curve at the point of contact. Do you know how to find an equation for the slope of a curve?
     
  6. Jul 21, 2017 #5

    scottdave

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    What expression would you use for m of the orthogonal line, if the parabola has a slope of 2x ? How can we take this information and create an equation for a line (yes m will seem to be variable). Now find the intersection point between that line, and the parabola.
     
  7. Jul 21, 2017 #6
    slope of the curve is 2x

    m for parabola = 2x
    m for the tangent line is 2x or 2?
    the m for orthogonal line is -1/(2x) or -1/2
     
  8. Jul 21, 2017 #7

    Mark44

    Staff: Mentor

    Thread moved. Questions involving derivatives belong in the Calculus & Beyond section, not the Precalc section.
     
  9. Jul 21, 2017 #8

    Ray Vickson

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    The OP is correct: he needs to know which tangent to use, because given any tangent you can find a unique line orthogonal to that tangent and passing through the point (0,3/2). However, if the intersection of the line and the tangent actually lies on the curve (in other words, if the line to be drawn passes both through (0,3/2) and the tangency point) then, indeed, there is a unique solution.
     
  10. Jul 21, 2017 #9

    scottdave

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    This is the way that I approached it. Let x = a on the parabola y = x2, so the point (a,a2) is on the parabola. The tangent line at that point has a slope of 2a, and the orthogonal line has a slope of -1/(2a). Draw a line from this point to the specified point (0,3/2) and solve for a.
     
  11. Jul 22, 2017 #10
    thanks, i think i get your point
     
    Last edited: Jul 22, 2017
  12. Jul 22, 2017 #11

    SammyS

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    Please give the complete statement of the problem as it was given to you.
     
  13. Jul 23, 2017 #12
    Consider a parabola y = x^2. Answer the following questions and fill in your responses in the corresponding boxes on the answer sheets

    (1) the line that goes through the point (0, 3/2) and is orthogonal to a tangent line to the part of parabola y = x^2 with x > 0 is y = Ax + 3/2
    Solve for A
     
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