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Find the equation of the normal to the graph at M(2,1)

  1. Jan 14, 2009 #1
    if:
    2x3+8lny+x3ey-16-8e=0

    what i did was
    dy/dx=[tex]\dot{y}[/tex]

    6x2+8(1/y)[tex]\dot{y}[/tex]+3x2ey+x3ey[tex]\dot{y}[/tex]=0

    [tex]\dot{y}[/tex]=-(x2(6+3ey)/(8/y+x3ey)

    then to find the incline of the normal, m=-1/[tex]\dot{y}[/tex]

    so at(2,1)

    [tex]\dot{y}[/tex]=-3(2+e)/2(1+e)

    m(N)=2(1+e)/3(2+e)

    does this seem right??

    Y-1=[2(1+e)/3(2+e)]{x-2}
     
  2. jcsd
  3. Jan 14, 2009 #2
    Yes, it's correct.
     
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