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Math Help - Equation of tangent line to following curve.

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    Equation of tangent line to following curve.

    Find the equation of the tangent lines to the following curve at the indicated point.
    y= x / y+a at (0,0)

    Thanks.
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    Quote Originally Posted by Paige05 View Post
    Find the equation of the tangent lines to the following curve at the indicated point.
    y= x / y+a at (0,0)

    Thanks.
    Is this
    y = \frac{x}{y + a} at (0, 0)?

    The solution method is to solve for the derivative at the point x = 0. This then is the slope of your tangent line, and you may use some version of the line equation and the given point to find the equation for the line.

    In this case we would probably want to use implicit differentiation:
    \frac{dy}{dx} = \frac{1(y + a) - x \cdot \frac{dy}{dx} }{(y + a)^2}

    So
    (y + a)^2 \frac{dy}{dx} = y + a - x \cdot \frac{dy}{dx}

     [(y + a)^2 + x] \frac{dy}{dx} = y + a

    \frac{dy}{dx} = \frac{y + a}{(y + a)^2 + x}

    At the point (0, 0) the value of the derivative is
    \frac{dy}{dx} = \frac{0 + a}{(0 + a)^2 + 0} = \frac{1}{a}

    So the slope of the tangent line at (0, 0) is 1/a. Thus the tangent line has the form:
    y = \frac{1}{a} \cdot x + b

    We know this line passes through the point (0, 0), so
    0 = \frac{1}{a} \cdot 0 + b

    Thus b = 0 and the tangent line is
    y = \frac{1}{a} \cdot x

    -Dan
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    Thank you Dan
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  4. #4
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    Hello, Paige05!

    Find the equation of the tangent lines (?) to the following curve at the indicated point.
    . . y \:= \:\frac{x}{y+a} at (0,0)

    We have: . y^2 + ay \:=\:x

    Differentiate implicitly: . 2yy' + ay' \:=\:1\quad\Rightarrow\quad y'(2y+a) \:=\:1\quad\Rightarrow\quad y' \:=\:\frac{1}{2y + a}

    At (0,0)\!:\;\;y' \:=\:\frac{1}{a}

    The equation of the tangent is: .  y - 0 \:=\:\frac{1}{a}(x-0)\quad\Rightarrow\quad\boxed{ y \:=\:\frac{1}{a}x}

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