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Finding tangent lines to a parabola.

  1. Oct 9, 2006 #1
    Using a graph to help you, find the equations of all lines through the origin tangent to the parabola
    [tex]y=x^2-2x+4[/tex].

    Sketch the lines on the graph.

    I'm not sure how to start this off, I know the the first derivative would give me the slope of the tangent line at any point. If I find that:
    [tex]y=x^2-2x+4[/tex]
    [tex]f'(x)=2x-2[/tex]

    Now, I know that my line has the slope [tex]2x-2[/tex] and it goes through the origin. So, I thought of plugging that into the slope-intercept form for a line, but it doesn't really work, I don't think:
    [tex]0=(2x-2)(0)+b[/tex]
    [tex]b=0[/tex]

    The [tex]x[/tex]'s go out, and I'm left with an equation of:
    [tex]y=(2x-2)x[/tex]
    [tex]y=2x^2-2x[/tex]

    When I graphed this parabola and the original one, I noticed that the graphs intersected at the points that the tangent lines would go through. So, to find those points, I set the equations equal to each other:
    [tex]x^2-2x+4=2x^2-2x[/tex]
    [tex]x^2-4=0[/tex]
    [tex](x-2)(x+2)=0[/tex]
    [tex]x=2,x=-2[/tex]

    Since I just got the x-values, I plugged them in to the original function to get the corresponding y-values:
    [tex]f(2)=(2)^2-2(2)+4=4[/tex]
    [tex]f(-2)=(-2)^2-2(-2)+4=12[/tex]

    Now, I can find the equations of the two tangent lines by using the points I found:
    First tangent line:
    (0, 0), (2, 4)
    [tex]\frac{4-0}{2-0}=\frac{4}{2}=2[/tex]
    [tex]y=2x[/tex]

    Second tangent line:
    (0, 0), (-2, 12)
    [tex]\frac{12-0}{-2-0}=\frac{12}{-2}=-6[/tex]
    [tex]y=-6x[/tex]

    Is this correct? It seems like there would be an easier way to go about doing this... Does anyone know?... It seems like I made things work in some very weird, rough way.
     
    Last edited: Oct 9, 2006
  2. jcsd
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