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

$x$-axis = $\frac{\sqrt{21}}{3}$, $\frac{-2\sqrt{21}}{3}$

I don't understand how to find the $y$-axis.

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Isn't the question asking you to find where the tangent to the curve $x^2+xy+y^2=7$ is horizontal and vertical? – Joe Johnson 126 Nov 14 '12 at 12:54
If so, you can use implicit differentiation. – Joe Johnson 126 Nov 14 '12 at 12:55
I tried that and got -2 sqrt 21/ 3, sqrt 21/ 3, which is the opposite of the x axis. Would that be correct? – Courtney Nov 14 '12 at 13:00
Note that our curve function is symmetric in $x$ and $y$. By symmetry, for parallel to $y$-axis we interchange the roles of $x$ and $y$. Thus (i) your computation is correct and (ii) you need not have computed. If we did not have symmetry, the $x$-axis argument could be imitated by finding $\frac{dx}{dy}$. – André Nicolas Nov 14 '12 at 13:04

Usually for these problems one uses the implicit derivative, which for this problem is $$y'=\frac{-2x-y}{x+2y}.$$ Then horizontal tangents occur when the top is zero, i.e. $2x+y=0$, and vertical tangents occur when the bottom is zero, i.e. $x+2y=0$. These equations are then plugged into the original relation $x^2+xy+y^2=7$ to get the actual coordinates of the points.

Note: Just noticed that Joe Johnson made this same suggestion re. implicit derivative!

EDIT: I got the horizontal tangents occur at $(x,y)=(+\sqrt{7/3},-2\sqrt{7/3})$ and at $(x,y)=(-\sqrt{7/3},+2\sqrt{7/3}).$ The vertical tangent points were like these, only switch the ordering of the pairs $(x,y)$. Maybe because the original ellipse $x^2+xy+y^2=7$ has its major axis at 45 degrees rotated.

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Thank you very much! It makes much more sense now! :) – Courtney Nov 14 '12 at 13:07

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