Stationary points of an implicit cubic

A curve CC has equation x3+y3−xy=Ax^3+y^3-xy=A, where AA is a non-zero constant.

(a) Show that any stationary points on CC lie on y=3x2y=3x^2. marks

(b) Find, in terms of AA, the xx-coordinates of the stationary points, and hence determine the range of values of AA for which CC has two distinct stationary points. marks

(c) Suppose that CC has two stationary points. Determine the nature of each of the stationary points found in part (b). marks

[2023/EJC/Prelim/II/4]

Total marks: 9

Solution

If this part uses an earlier result, revisit (b) coordinates Quadratics in a power or open all working together.

(b) range of A

Discriminant

Hence determine the range of values of AA for which CC has two distinct stationary points.

Explore this idea

Hint

  1. Two distinct values of x3x^3 need a positive discriminant: 1+27A>01+27A>0.

Answer · (b) range of A

A>−1/27A>-1/27, with A≠0A\ne0.

At A=−1/27 the repeated candidate is singular, not a regular stationary point.

Explore the idea

Open interactive exploration

How many x-values give 27x⁶ − 2x³ = A?

Each intersection supplies an algebraic stationary candidate. The question excludes A = 0.

Height of the horizontal linek = 0.1
Try an example
Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct stationary candidates
2

Two stationary candidates for the nonzero parameter A.

All listed intersections are within the original drawing window. A touching point counts once; hollow endpoints are excluded.

Coordinates are rounded to 6 significant figures. Inspect a row to highlight that intersection and bring it into view.
Pointxy
-0.3246110.1
0.476630.1
  • 27x⁶ − 2x³
  • Y = A
  • X-coordinates of stationary points
  • Intersection selected in the coordinate table
Key idea

The sextic's minimum value is −127-\frac1{27} (at x=13x=\frac13): below that there are none.

Ideas to try
Find the lowest AA that still gives a stationary point.