Find the points on the curve $x^2+xy+y^2=7$ where tangent is parallel to (a) X axis (b) parallel to Y axis.

For part (a) if we differentiate w.r.t $x$ we get $y'=\large-\frac{2x+y}{x+2y}$, this when equated to zero (parallel to X axis means slope is zero) gives $x=-y/2$, this when substituted in the equation of the curve gives $\left(\sqrt{\frac{7}{3}},-2\sqrt{\frac{7}{3}}\right),\left(-\sqrt{\frac{7}{3}},2\sqrt{\frac{7}{3}}\right)$

For part (b) we find $dx/dy$, equate it to zero and find that the points are $\left(-2\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\right),\left(2\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}\right)$

$\underline{\text{My question}}$ is if we consider two points on the curve $(x_1, y_1),(x_2, y_2)$ such that $x_1 = x_2$, because for the line parallel to Y axis coordinates of x remain the same. Substituting these two points in the equation of the curve we get

$$x_1^2+x_1y_1+y_1^2=7$$ $$x_1^2+x_1y_2+y_2^2=7$$

solving them simultaneously we get $x_1=-(y_1+y_2)=x_2$, but this does not agree with the points that we calculated $\left(-2\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\right),\left(2\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}\right)$ in part (b). $y_1+y_2=0\neq x_1 \hspace{10pt}\text{OR} \hspace{10pt} \neq x_2$

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    $\begingroup$ WolframAlpha gets $y_1 = \frac12\left(\sqrt{28-3x^2} - x\right)$ and $y_2 = \frac12\left(-\sqrt{28-3x^2} - x\right)$. So we do have $y_1+y_2 = -x_1$, which when we move to a single point $(x, y)$ of tangency becomes $x = -2y$, which is true for both of your solutions. But $y_1+y_2 \neq 0$. $\endgroup$ – Arthur Oct 27 '16 at 5:29

If $x_1,x_2$ with $x_1=x_2$ are the roots of the equation $$x^2+xy+y^2-7=0$$

we need the discriminant$=0$ $$\implies y^2-4(y^2-7)=0\iff y^2=?$$ and consequently $$x_1=x_2=-\dfrac y2$$


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