Locus of Point $x^3+y^3= 8- 6xy$

[kaliprasad]

find locus of point $x^3+y^3= 8- 6xy$

 

[chisigma]

find locus of point $x^3+y^3= 8- 6xy$

[sp]The solutions of the equation in y ...

$\displaystyle y^{3} + 6\ x\ y + x^{3} - 8 =0\ (1)$

... are...

$\displaystyle y = 2 - x$

$\displaystyle y = \frac{1}{2}\ \{ x - 2 - i\ \sqrt{3}\ (x + 2)\}$

$\displaystyle y = \frac{1}{2}\ \{ x - 2 + i\ \sqrt{3}\ (x + 2)\}\ (2)$

... so that in the real x-y plane the locus is the straight line $\displaystyle y = 2 - x$...[/sp]

Kind regards

$\chi$ $\sigma$

 

[kaliprasad]

[sp]The solutions of the equation in y ...

$\displaystyle y^{3} + 6\ x\ y + x^{3} + 8 =0\ (1)$

... are...

$\displaystyle y = 2 - x$

$\displaystyle y = \frac{1}{2}\ \{ x - 2 - i\ \sqrt{3}\ (x + 2)\}$

$\displaystyle y = \frac{1}{2}\ \{ x - 2 + i\ \sqrt{3}\ (x + 2)\}\ (2)$

... so that in the real x-y plane the locus is the straight line $\displaystyle y = 2 - x$...[/sp]

Kind regards

$\chi$ $\sigma$

you have done a mistake(typo error) in 1st line and one more solution is missing

 

[MarkFL]

My solution:

Spoiler

The first thing I notice is that there is cyclic symmetry between $x$ and $y$, and so setting $y=x$, we find:

\(\displaystyle x^3+6x^2+x^3=2x^3+6x^2=8\)

\(\displaystyle x^2(x+3)-4=0\)

\(\displaystyle (x-1)(x+2)^2=0\)

Thus, we know the points:

\(\displaystyle (-2,-2),\,(1,1)\)

are on the given curve. Next, if we begin with the line:

\(\displaystyle y=2-x\)

and cube both sides:

\(\displaystyle y^3=8-12x+6x^2-x^3\)

We may arrange this as:

\(\displaystyle x^3+3\cdot2x(2-x)+y^3=8\)

Since $y=2-x$, we obtain:

\(\displaystyle x^3+6xy+y^3=8\)

Now since the point $(1,1)$ is on the line $y=2-x$ we conclude the locus of points is the line:

\(\displaystyle y=2-x\) and the point $(-2,-2)$.

 

[Opalg]

Having seen previous answers makes this easier!
[sp]$x^3 - 6xy + y^3 - 8 = (x+y-2)(x^2 - xy + y^2 + 2x + 2y + 4)$, so the locus consists of the line $x+y-2=0$ together with that given by $0 = x^2 - xy + y^2 + 2x + 2y + 4 = \frac14\bigl((x + y + 4)^2 + 3(x-y)^2\bigr).$ But then $x+y+4 = x-y = 0$, which has the unique solution $(x,y) = (-2,-2).$ So the locus consists of that line together with that point.[/sp]

 

[kaliprasad]

[sp]The solutions of the equation in y ...

$\displaystyle y^{3} + 6\ x\ y + x^{3} - 8 =0\ (1)$

... are...

$\displaystyle y = 2 - x$

$\displaystyle y = \frac{1}{2}\ \{ x - 2 - i\ \sqrt{3}\ (x + 2)\}$

$\displaystyle y = \frac{1}{2}\ \{ x - 2 + i\ \sqrt{3}\ (x + 2)\}\ (2)$

... so that in the real x-y plane the locus is the straight line $\displaystyle y = 2 - x$...[/sp]

Kind regards

$\chi$ $\sigma$

Good ans by Mark and Oplag

Spoiler

in the above from the 2nd and 3rd line the imaginary part could have been set to zero to get x = y = -2

My solution

Spoiler

take 8 to the left

$x^3+y^3-8 + 6xy=0$

or $x^3+y^3 + (-2)^3 - 3(-2) xy= 0$
above is of the form $a^3+b^3+c^3-3abc=0$

so $(x+y-2)\dfrac{1}{2}((x-y)^2 + (x+2)^2+(y+2)^2)=0$

comparing we get $x = y = -2$ or $x + y- 2 = 0$

so the locus is straight line $x+y=2$ and point (-2,-2)

 

[MarkFL]

This problem was very similar to a problem posted somewhere (I don't recall where) as a challenge, and I relied on my work from that to answer this question. Here is the other question:

Spoiler

Show that the curve $x^3+3xy+y^3=1$ has only one set of three distinct points, $P$, $Q$, and $R$ which are the vertices of an equilateral triangle, and find its area.

My solution:

The first thing I notice is that there is cyclic symmetry between $x$ and $y$, and so setting $y=x$, we find:

\(\displaystyle 2x^3+3x^2-1=(x+1)^2(2x-1)=0\)

Thus, we know the points:

\(\displaystyle (x,y)=(-1,-1),\,\left(\frac{1}{2},\frac{1}{2} \right)\)

are on the given curve. Next, if we begin with the line:

\(\displaystyle y=1-x\)

and cube both sides, we obtain:

\(\displaystyle y^3=1-3x+3x^2-x^3\)

We may arrange this as:

\(\displaystyle x^3+3x(1-x)+y^3=1\)

Since $y=1-x$, we may now write

\(\displaystyle x^3+3xy+y^3=1\)

And since the point \(\displaystyle \left(\frac{1}{2},\frac{1}{2} \right)\) is on the line $y=1-x$, we know the locus of the given curve is the line $y=1-x$ and the point $(-1,-1)$. Hence, there can only be one set of points on the given curve that are the vertices of any triangle, equilateral or otherwise.

Using the formula for the distance between a point and a line, we find the altitude of the equilateral triangle will be:

\(\displaystyle h=\frac{|(-1)(-1)+1-(-1)|}{\sqrt{(-1)^2+1}}=\frac{3}{\sqrt{2}}\)

Using the Pythagorean theorem, we find that the side lengths of the triangle must be:

\(\displaystyle s=\frac{2}{\sqrt{3}}h=\sqrt{6}\)

And so the area of the triangle is:

\(\displaystyle A=\frac{1}{2}sh=\frac{1}{2}\sqrt{6}\frac{3}{\sqrt{2}}=\frac{3\sqrt{3}}{2}\)