0
$\begingroup$

"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.

$\endgroup$
2
$\begingroup$

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 $(a, -2a)$.

$\endgroup$
0
$\begingroup$

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 $(a,-2a)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.