Identify a parametric self-intersection · source reference

Printed question

A curve $C$ has parametric equation $x=t^2-1$, $y=4t-t^3+15$, where $t\in\mathbb R$.

(i) Show that the curve $C$ cuts the $x$-axis at the point $(8,0)$ and the $y$-axis at $(0,12)$ and $(0,18)$. [3]

(ii) Sketch $C$, giving the equation of the line of symmetry. [2]

(iii) Find the exact coordinates of the point where $C$ crosses itself. [2]

[2020/NYJC/Promo/10]

Printed working

(i) Show that the curve $C$ cuts the $x$-axis at $(8,0)$. Let $y=0$.

$y=4t-t^3+15=0\Rightarrow t=3$. Thus $x=3^2-1=8$.

Hence, the curve $C$ cuts the $x$-axis at $(8,0)$.

To find the $y$-intercepts, we let $x=0$, then

$x=t^2-1=0\Rightarrow t=1,-1$.

When $t=1$, $y=4t-t^3+15=4-1+15=18$.

When $t=-1$, $y=4t-t^3+15=-4+1+15=12$.

Hence, the $y$-intercepts are $(0,18),(0,12)$.

(ii) Printed calculator graph below.

(iii) Observe that the point where $C$ crosses itself lies on the symmetric axis with equation $y=15$, thus its $y$-ordinate is 15.

Substitute $y=15$ into $y=4t-t^3+15$,

$15=4t-t^3+15\Rightarrow t(4-t^2)=0\Rightarrow t=0,2,-2$.

When $t=0$, $x=t^2-1=-1,y=15$, $(-1,15)$ is not the required point.

When $t=\pm2$, $x=t^2-1\Rightarrow x=(\pm2)^2-1=3$.

Thus, the curve crosses itself at $(3,15)$.

Teacher notes

Using GC to sketch curve:

1. Go to “mode”. Set to “radian”, “parametric”.
2. Go to “y=”. Key in eqn of curve.
3. Go to “window”. Change $T_{\min}$ to a negative value (e.g. −10) and reset the ranges of $x,y$ (e.g. $X_{\min}=-5,X_{\max}=15,Y_{\min}=-5,Y_{\max}=30$).

Printed answer key

(ii) Line of Symmetry: y=15; native graph above. (iii) (3,15). The short key omits (i).

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: GRAPHS AND CURVE SKETCHING. Question/answer booklet printed pages 1–14; solutions printed pages 1–28. Running metadata and repeated headings are transcribed in the topic metadata article.