Given the curve x²-xy+y²-1=0. 1° Show that there exist

Given the curve $x^{2}-xy+y^{2}-1=0$. 1° Show that there exist two distinct points on the curve where the tangent is parallel to a given line. Can an arbitrary line be used? 2° At which points of the curve must normals be drawn so that they pass through the origin?

Step 1: Implicit differentiation

With $F(x,y)=x^{2}-xy+y^{2}-1$: \[2x-y-x\,y'+2y\,y' = 0 \;\Longrightarrow\; y' = \frac{y-2x}{2y-x}, \qquad x\neq2y. \tag{1}\]

Step 2: Part 1°: tangents parallel to a given line

Let the given line have slope $m$. Setting $y'=m$ in (1) and solving for $y$ in terms of $x$: \[y-2x = m(2y-x) \;\Longrightarrow\; y(1-2m) = x(2-m) \;\Longrightarrow\; y = \lambda x, \qquad \lambda=\frac{2-m}{1-2m}\ \ (m\neq\tfrac12). \tag{2}\]

Substituting $y=\lambda x$ into the curve equation, $x^{2}(1-\lambda+\lambda^{2})=1$, so

\[ x = \pm\frac{1}{\sqrt{1-\lambda+\lambda^{2}}}. \tag{3} \]

Since $1-\lambda+\lambda^{2}=(\lambda-\tfrac12)^{2}+\tfrac34>0$ for every real $\lambda$, equation (3) always yields two real, distinct points $P_{1}=(x_{1},\lambda x_{1})$ and $P_{2}=(-x_{1},-\lambda x_{1})$.

Special case $m=\tfrac12$: (2) gives $x=0$, $y=\pm1$ — again two points. Vertical tangents ($2y=x$): substituting into the curve gives $3y^{2}=1$, $y=\pm1/\sqrt3$ — again two points.

\[\text{For every direction (including vertical), there are exactly two points with parallel tangents.}\]

This works for any line because the curve $x^{2}-xy+y^{2}=1$ is a centrally symmetric (tilted) ellipse: every tangent direction occurs at two antipodal points. Not every curve has this property, but this one does.

Step 3: Part 2°: normals through the origin

The normal at $(x_{0},y_{0})$ has slope $-1/y'=-\dfrac{2y_{0}-x_{0}}{y_{0}-2x_{0}}$; requiring it to pass through $(0,0)$ and cross-multiplying, \[-y_{0}(y_{0}-2x_{0}) = x_{0}(2y_{0}-x_{0}) \;\Longrightarrow\; x_{0}^{2}=y_{0}^{2} \;\Longrightarrow\; y_{0}=\pm x_{0}. \tag{4}\]

Case $y_{0}=x_{0}$: the curve gives $x_{0}^{2}=1$, so $(1,1)$ and $(-1,-1)$.

Case $y_{0}=-x_{0}$: the curve gives $3x_{0}^{2}=1$, so $\left(\tfrac1{\sqrt3},-\tfrac1{\sqrt3}\right)$ and $\left(-\tfrac1{\sqrt3},\tfrac1{\sqrt3}\right)$.

Step 4: Verification

\[\text{All four points on the curve: } 1-1+1-1=0\ \checkmark;\quad 3\cdot\tfrac13+\tfrac13+\tfrac13-1=0\ \checkmark\]

\[\text{Normal at } (1,1)\text{: } y'=\tfrac{1-2}{2-1}=-1,\ \text{normal slope }1,\ y=x,\ \text{through } O\ \checkmark\]

\[\text{Normal at } \left(\tfrac1{\sqrt3},-\tfrac1{\sqrt3}\right)\text{: } y'=1,\ \text{normal slope }-1,\ y=-x,\ \text{through } O\ \checkmark\]