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  1. $x^2+xy+y^2=7$
  2. Find $dy/dx$
  3. $dy/dx= (-2x-y)/(x+2y)$

How do I take $dy/dx$ and get the equation of the tangent line parallel to the $x$-axis?

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What's your question, specifically? – Marra Sep 28 '12 at 1:06
Combine (1) with $\frac{dy}{dx} = 0$ from (3). If you have done it correctly, you get two lines, since (1) gives an ellipse centered at the origin. – Will Jagy Sep 28 '12 at 1:06
The question is "what is the equation of the tangent line of x^2+xy+y^2=7 parallel to the x-axis?" – Paul Sep 28 '12 at 1:13
The tangent line occurs when $\dfrac{dy}{dx}(x)=0$. As Will pointed out you have everything needed to find it. – Marra Sep 28 '12 at 1:15

You have

$$ x^2 + xy + y^2 = 7. $$

Using implicit differentiation you get $$ \begin{align} \frac{d}{dx} x^2 + xy + y^2 &= \frac{d}{dx} 7 &\Rightarrow\\ 2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} &= 0. \end{align} $$

From this you solve for $\frac{dy}{dx}$. You get [you can do this] $$ \frac{dy}{dx} = \frac{-2x-y}{x + 2y} $$ To find the equation of the tangent line that is parallel to the $x$-axis, you need to determine the $x$ and $y$ such that the numerator is zero and the denominator is not zero. At those points you have a slope of zero. Then you can simply write down the equaltion of the tangent line (the slope is obviously zero, so...).

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is the answer y=sqrt(7)? – Paul Sep 28 '12 at 1:43
@Paul / Matt: So if $y = -2x$ then from the original equation you have $3x^2 = 7$, so ... – Thomas Sep 28 '12 at 1:53
A small matter of my idiosyncratic preference. From $2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0$, I would immediately set $\frac{dy}{dx}=0$, getting $2x+y=0$. Division makes me nervous. – André Nicolas Sep 28 '12 at 2:58
@AndréNicolas: That would be much better! – Thomas Sep 28 '12 at 12:50

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