Finding stationary point of an implicit equation: $ xy(x+y)=2a^3 $

Question

"The equation of a curve is $$ xy(x+y)=2a^3 $$ where $a$ is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the $x$-axis, and find the co-ordinates of this point."

The above question came up in A Level P3 pastpapers. I was able to simplify it to this point: $$ xy(x+y)=2a^3 $$ $$ x^2y+xy^2-2a^3=0 $$ Then differentiation: $$ x^2{dy\over dx}+2xy+2xy{dy\over dx}+y^2=0 $$ $$ {dy\over dx}(x^2+2xy)+2xy+y^2=0 $$ $$ {dy\over dx}=-{y(y+2x)\over x(x+2y)} $$ I don't know what to do after this step to get the solution. Please help.

Answer 1

Call the point $(x_0, y_0)$. Since its tangent is parallel to the $x$-axis, its derivative there is $0$, so we know that either $y_0 = 0$ or $(y_0 + 2x_0) = 0$.

But we know that $y_0 \neq 0$. Otherwise, plugging it into the curve would imply that $a = 0$, a contradiction. Hence, we know that $y_0 = -2x_0$. Substituting this into the curve, we have: $$ x_0(-2x_0)(x_0 + (-2x_0)) = 2a^3 \iff 2x_0^3 = 2a^3 \iff x_0 = a $$ So the point is $(a, -2a)$.

Answer 2

Let F(x,y) = $xy(x+y) - 2a^3$

We need the gradient of F(x,y), which we can find as,

Differentiating F(x,y) w.r.t to x yields $2xy+y^2$
Differentiating F(x,y) w.r.t y yields $x^2 + 2xy$
For a stationary point, both must be equal to 0, so it can be easily seen that the only solution is (x,y) = (0,0,)

Edit: Since a>0, this function does not have a stationary point - (0,0) does not lie on the function.
However since you want to find a point at where the tangent to the function is parallel to the x axis, set the Gradient of F(x,y) w.r.t x =0 yielding $2xy + y^2$, so either $y=0$ or $y = -2x$, and substituting this back into the original equation yields x, from which you can get y.
As Adriano has pointed out the solution will be $(a,-2a)$