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Finding 2 values

  1. Sep 27, 2009 #1
    For what values of a and b is the line 4x + y = b tangent to the parabola y = ax2 when x = 3?



    4x + y = b
    y = ax2




    Tried making 4x + y = b to y = b - 12 and making that equal y = 9a then making and setting one of those values to 0. But that was winging it with no prior knowledge. Can someone help me please?
     
  2. jcsd
  3. Sep 27, 2009 #2
    Hi xpack, well you were sort of (ish :D) starting in the write direction, but I think Ill try to help you from the beginning. So with any question its always important to dissect it before even trying to start it. First ill number the equations so we can refer to them by numbers:

    [tex]
    (i) 4x + y = b [/tex]
    [tex](ii) y = ax^2
    [/tex]

    So in we are looking for the tangent to curve (ii), so immediately that should be ringing alarm bells with you "Differentiation", we need to find the gradient of (ii).

    Now you should be familiar with the equation:

    [tex]
    y - y_1 = m(x - x_1)
    [/tex]

    in case you not this is the general form that any line in a 2 dimensional space can take, where y and x are our variables and x1 and y1 represent the coordinates of a known point on the line. This can be rearranged to:

    [tex]
    (iii)y = mx + (y_1 - mx_1)
    [/tex]

    which you should see is in the form y = mx + c, which defiantly should be familiar to you. Now know this, and the fact that you have just found the differential (gradient) of the parabola can you do some sort of comparison to create your own equation, what must be the same about (iii) and your differential of the parabola, from this you should be able find a.

    Now if you can get that far then you should be able to do a similar thing with the (y1 - mx1) part, what does this equal in the equations you gave (hint look at (i) ;D). This one is a little harder than the first part but isnt too tricky,

    Hope that helps xpack :D
     
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