Stationary points of an implicit cubic · source reference

Printed question

A curve $C$ has equation $x^3+y^3-xy=A$, where $A$ is a non-zero constant.

(a) Show that any stationary points on $C$ lie on $y=3x^2$. [2]

(b) Find, in terms of $A$, the $x$-coordinates of the stationary points, and hence determine the range of values of $A$ for which $C$ has two distinct stationary points. [4]

(c) Suppose that $C$ has two stationary points. Determine the nature of each of the stationary points found in part (b). [3]

[2023/EJC/Prelim/II/4]

Printed working

(a) Differentiate $x^3+y^3-xy=A$:

$$3x^2+3y^2\frac{dy}{dx}-x\frac{dy}{dx}-y=0\Rightarrow(3y^2-x)\frac{dy}{dx}=-3x^2+y\quad(*).$$

When $dy/dx=0$, $0=-3x^2+y\Rightarrow y=3x^2$ (shown).

(b) Substitute $y=3x^2$:

$$x^3+(3x^2)^3-x(3x^2)=A\Rightarrow x^3+27x^6-3x^3=A\Rightarrow27x^6-2x^3-A=0.$$ $$x^3=\frac{-(-2)\pm\sqrt{(-2)^2-4(27)(-A)}}{2(27)}=\frac{2\pm2\sqrt{1+27A}}{2(27)}=\frac{1\pm\sqrt{1+27A}}{27},$$ $$x=\sqrt[3]{\frac{1\pm\sqrt{1+27A}}{27}}=\frac13\sqrt[3]{1\pm\sqrt{1+27A}}.$$

For more than one stationary point, discriminant $1+27A>0$, so $A>-1/27$, retaining the given $A\ne0$.

(c) Differentiate (*):

$$(3y^2-x)\frac{d^2y}{dx^2}+\left(6y\frac{dy}{dx}-1\right)\frac{dy}{dx}=-6x+\frac{dy}{dx}.$$

At a stationary point $dy/dx=0$, $y=3x^2$:

$$[3(3x^2)^2-x]\frac{d^2y}{dx^2}=-6x\Rightarrow(27x^4-x)\frac{d^2y}{dx^2}=-6x,$$ $$\frac{d^2y}{dx^2}=-\frac{6x}{27x^4-x}=-\frac6{27x^3-1}.$$

When $x^3=(1+\sqrt{1+27A})/27$, $y''=-6/\sqrt{1+27A}<0$: maximum.

When $x^3=(1-\sqrt{1+27A})/27$, $y''=6/\sqrt{1+27A}>0$: minimum.

Teacher notes

For any stationary point on $C$, $dy/dx=0$.

Answer key

(b) $x=\frac13\sqrt[3]{1\pm\sqrt{1+27A}}$, $A>-1/27$ (and given $A\ne0$).