Show that the graph of the equation $x^{3}+3 x^{2} y+3 x y^{2}+y^{3}-x^{2}+y^{2}=0$ is a union of line and a parabola.

Question

Show that the graph of the equation $$x^{3}+3 x^{2} y+3 x y^{2}+y^{3}-x^{2}+y^{2}=0$$ is a union of line and a parabola.

This seems a pretty easy question, but the answer I got is not matching with the claim made in the question.

Here's how I solved the problem:

The idenitity $(x+y)^3=x^3+y^3+3xy(x+y)$ makes the problem easy. We then have, $(x+y)^3-x^2+y^2=0.$ Now, if $x+y=1$, then we can write, $(x+y)^3-x^2+y^2=0,$ implies $1-x^2+y^2=0$ which means, $x^2-y^2=1$ and this is the equation of a rectangular hyperbola and $x+y=1$ is a straight line equation.

But I am unable to find how this equation is the union of a line and a parabola. It would be much helpful if someone points out where is the mistake in my solution.

Answer 1

We have

$$\color{blue}{x^{3}+3 x^{2} y+3 x y^{2}+y^{3}}-x^{2}+y^{2}=\color{blue}{(x+y)^3}-(x+y)(x-y)=$$

$$=(x+y)\left[\color{blue}{(x+y)^2}+(x-y)\right]=0$$

and with respect to the orthogonal directions $u=x+y=0$ and $v=x-y=0$

$$(x+y)^2+(x-y)=0 \iff v=-u^2$$

which is indeed a parabola.

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The mistake you are doing is that you are just plugging $x+y=1$ into the original equation which is another thing (i.e. an intersection).

Here we are requested to factor the given expression $f(x,y)=l(x,y)\cdot p(x,y)=0$ such that $l(x,y)=0$ is a line and $p(x,y)=0$ is a parabola, that is $$f(x,y)=l(x,y)\cdot p(x,y)=0 \iff l(x,y)=0 \; \lor\; p(x,y)=0$$

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Here is a plot by WA

Answer 2

Just to show the absurdity of what you did: Suppose you have the equation $x+y=2$ instead and you substitute $x+y=1$ in your equation. Then you get $1=2$ which is absurd.

Furthermore, if you substitute $x+y=1$ in your equation then you have to substitute also $y=1-x$ getting $2-2x=0.$ So, $x=1$ and $y=0$. That means $(1,0)$ is the intersection of the graph of your equation $(x+y)^3-x^2+y^2=0$ and the line $x+y=1.$