Number of right isosceles triangles that can be formed with points lying on the curve $8x^3+y^3+6xy=1$

Question

Number of right isosceles triangles that can be formed with points lying on the curve $$8x^3+y^3+6xy=1$$

MY ATTEMPT : We have, $$8x^3+y^3+6xy=1$$ adding both the sides $$6xy^2+12x^2$$ and simplifying we get ,
$$y^2+y(1-2x)+4x^2+2x+1=0$$ after this I got struck pls help me out with this question

Answer given is 3

Answer 1

Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we see that our equation it's $$(2x+y-1)(4x^2+y^2+1-2xy+2x+y)=0,$$ which gives $$2x+y-1=0$$ or $$(2x-y)^2+(2x+1)^2+(y+1)^2=0.$$ The second gives a point $A\left(-\frac{1}{2},-1\right)$ and two other points are placed on the line $2x+y-1=0$.

Thus, the vertex of the right angle of the triangle may be in three these points.

Easy to see that all these possibilities occur.

Indeed, let $AD$ be a perpendicular to the line $l:2x+y-1=0$ and $\{B,C\}\subset l$ such that $D$ be a mid-point of $BC$ and $AD=DB=DC.$

Thus, $\Delta ABC$, $\Delta ADB$ and $\Delta ADC$ they are right-angled isosceles triangles.