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Thread: Normals parallel to a line

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    Normals parallel to a line

    My brother and I have been working on this problem for the better part of the day. I think we keep overlooking something. Please help us with this problem.

    Normals parallel to a line: Find the normals to the curve xy+2x-y=0 that are parallel to the line 2x+y=0.

    Thank you in advance for your responces.
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    Quote Originally Posted by AnDiesel View Post
    My brother and I have been working on this problem for the better part of the day. I think we keep overlooking something. Please help us with this problem.

    Normals parallel to a line: Find the normals to the curve xy+2x-y=0 that are parallel to the line 2x+y=0.

    Thank you in advance for your responces.
    $\displaystyle xy + 2x - y = 0$

    $\displaystyle (x - 1)y = -2x$

    $\displaystyle y = -\frac{2x}{x - 1}$

    Now find the slope of the line tangent to this curve at (x, y):
    $\displaystyle y' = \frac{2}{(x - 1)^2}$

    The normal to the tangent line has the slope
    $\displaystyle -\frac{1}{y'} = -\frac{(x - 1)^2}{2} = -\frac{1}{2}x^2 + x - \frac{1}{2}$

    So given a point (x, y) on the original curve you now know the slope of the line normal to the curve at (x, y). I'll let you finish it out from here.

    -Dan
    Last edited by topsquark; Jul 3rd 2008 at 05:08 AM.
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    Ok, I see what you did. I kept differentiating implicity and I must have been overlooking a mistake (even though it should have worked too). But I came up with y=-2x+3 , y=-2x-3 as the equations of my normals. Everything looks good when I graph it too, so I think this is right. Thank you very much for your quick responce and your help.
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    Quote Originally Posted by topsquark View Post
    $\displaystyle xy {\color{red}+} 2x - y = 0$ Mr F says: Dang it!!

    $\displaystyle (x - 1)y = 2x$

    $\displaystyle y = \frac{2x}{x - 1}$

    Now find the slope of the line tangent to this curve at (x, y):
    $\displaystyle y' = \frac{-2}{(x - 1)^2}$

    The normal to the tangent line has the slope
    $\displaystyle -\frac{1}{y'} = -\frac{(x - 1)^2}{-2} = \frac{1}{2}x^2 - x + \frac{1}{2}$

    So given a point (x, y) on the original curve you now know the slope of the line normal to the curve at (x, y). I'll let you finish it out from here.

    -Dan
    There's a small mistake (edited in red - dang it, the very first line. Topsquawk was obviously thinking about Dora at the time .... lol!!). There's obviously a flow-on effect but the corrections are simple.

    You should get the x-coordinates of the required points on the given curve are x = 3 and x = -1.

    Edit: Reckoner, you should have posted your reply!
    Last edited by mr fantastic; Jul 2nd 2008 at 06:37 PM.
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    Quote Originally Posted by AnDiesel View Post
    Ok, I see what you did. I kept differentiating implicity and I must have been overlooking a mistake (even though it should have worked too). But I came up with y=-2x+3 , y=-2x-3 as the equations of my normals. Everything looks good when I graph it too, so I think this is right. Thank you very much for your quick responce and your help.
    Obviously there's a bit more work if you use implicit differentiation (and therefore don't make y the subject). Anyway, since you mentioned it:

    Implicit differentiation:

    $\displaystyle y + x \frac{dy}{dx} + 2 - \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} \, (x - 1) + y + 2 = 0 \Rightarrow \frac{dy}{dx} = - \frac{(y +2)}{(x - 1)}$.

    Therefore you require $\displaystyle -2 = \frac{x-1}{y+2} \Rightarrow -2y - 4 = x - 1$

    $\displaystyle \Rightarrow x + 2y = -3$ ..... (1)

    Since the points on the curve that the required normals pass through obviously lie on the curve, you also require

    $\displaystyle xy + 2x - y = 0$ ...... (2)

    Solve equations (1) and (2) simultaneously:

    x = 3 and y = -3 or x = -1 and y = -1.


    Equation of normal 1: m = -2 and passes through the point (3, -3). $\displaystyle y - (-3) = -2(x - 3) \Rightarrow y = -2x + 3$.

    Equation of normal 2: m = -2 and passes through the point (-1, -1). $\displaystyle y - (-1) = -2(x - (-1)) \Rightarrow y = -2x - 3$.
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    Thank you for your replies; they were a great help to me.
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    Quote Originally Posted by mr fantastic View Post
    There's a small mistake (edited in red - dang it, the very first line. Topsquawk was obviously thinking about Dora at the time .... lol!!). There's obviously a flow-on effect but the corrections are simple.

    You should get the x-coordinates of the required points on the given curve are x = 3 and x = -1.

    Edit: Reckoner, you should have posted your reply!
    Thanks for pointing out the mistake. (Not for the Dora comment!) I have corrected my original post.

    -Dan
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