Review question

# Can we show these two cubic curves touch? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8543

## Solution

A curve is given by the equation $\begin{equation*} y = ax^3 - 6ax^2 + (12a + 12)x - (8a + 16), \label{eq:1repeat}\tag{*} \end{equation*}$ where $a$ is a real number. Show that this curve touches the curve with equation $\begin{equation*} y = x^3 \label{eq:2repeat}\tag{*{*}} \end{equation*}$

at $(2,8)$. Determine the coordinates of any other point of intersection of the two curves.

Let $f(x) = ax^3 - 6ax^2 + (12a + 12)x - (8a + 16)$ and $g(x) = x^3.$

To find the points of intersection, we set $f(x) = g(x)$ to get \begin{align*} & ax^3 - 6ax^2 + (12a + 12)x - (8a + 16) = x^3 \\ \iff & (a - 1)x^3 - 6ax^2 + (12a + 12)x - (8a + 16) = 0 \\ \iff & (x - 2)[(a - 1)x^2 - (4a + 2)x + (4a + 8)] = 0 \\ \iff & (x - 2)^2[(a - 1)x - (2a + 4)] = 0. \end{align*}

So the points of intersection have coordinates (2,8) and $\left(\frac{2a+4}{a-1},\left[\frac{2a+4}{a-1}\right]^3\right).$

The double root for $x = 2$ means that we have a touching there; the two curves are tangent to each other at $(2,8)$.

Alternatively…

To show that the curves $\eqref{eq:1repeat}$ and $\eqref{eq:2repeat}$ touch at $(2,8)$, we need to show that both curves pass through $(2,8)$ and that the gradients of the curves are equal at this point.

Now $f(2) = 8a - 24a + 2(12a + 12) - (8a + 16) = 8$ and $g(2) = 8,$ so both curves pass through the point $(2,8)$.

Also $f'(2) = 12a - 24a + (12a + 12) = 12$ and $g'(2) = 12,$ so the curves have the same gradient at the point $(2,8)$.

Therefore the curves $\eqref{eq:1repeat}$ and $\eqref{eq:2repeat}$ touch at $(2,8)$.

1. Sketch on the same axes the curves $\eqref{eq:1repeat}$ and $\eqref{eq:2repeat}$ when $a = 2$.

When $a = 2$, $\eqref{eq:1repeat}$ becomes $y = 2x^3 - 12x^2 + 36x - 32.$ From above, we have that the two graphs intersect at $(8,512)$ and touch at $(2,8)$.

Now differentiating gives $\frac{dy}{dx} = 6x^2 - 24x + 36,$ so stationary points occur when \begin{align*} & 6x^2 - 24x + 36 = 0 \\ \iff & x^2 - 4x + 6 = 0 \\ \iff & (x - 2)^2 = -2. \end{align*}

This equation has no real solutions, so there are no stationary points.

Further, we can see from this equation that $y'>0$ for all $x$, and so $y$ is an increasing function.

Since $y<0$ when $x=1$ but $y>0$ when $x=2$, we know that it has a root (which must be the only root since $y$ is increasing) somewhere between $1$ and $2$.

We also know that $\eqref{eq:1repeat}$ with $a = 2$ intersects the $y$ axis at $(0,-32)$.

So when $a = 2$, the graphs must have the following shape.

1. Sketch on the same axes the curves $\eqref{eq:1repeat}$ and $\eqref{eq:2repeat}$ when $a = 1$.

When $a = 1$, $\eqref{eq:1repeat}$ becomes $y = x^3 - 6x^2 + 24x - 24.$ The two graphs touch at $(2,8)$ but do not intersect elsewhere because $\dfrac{2a+4}{a-1}$ is undefined when $a = 1$.

Differentiating gives $\frac{dy}{dx} = 3x^2 - 12x + 24,$ so stationary points occur when \begin{align*} & 3x^2 - 12x + 24 =0 \\ \iff & x^2 - 4x + 8 = 0 \\ \iff & (x - 2)^2 = -4. \end{align*}

This equation has no real solutions, so there are no stationary points.

Again, this means that $y'>0$ for all $x$, and so $y$ is an increasing function.

Since $y<0$ at $x=1$ but $y>0$ at $x=2$, we know that the graph of $y$ crosses the $x$-axis somewhere between $x=1$ and $x=2$, which should help our sketch.

We also know that $\eqref{eq:1repeat}$ with $a = 1$ intersects the $y$ axis at $(0,-24)$.

So when $a=1$, the graphs have the following shape.

1. Sketch on the same axes the curves $\eqref{eq:1repeat}$ and $\eqref{eq:2repeat}$ when $a = -2$.

When $a = -2$, $\eqref{eq:1repeat}$ becomes $y = -2x^3 + 12x^2 - 12x=-2x(x^2-6x+6).$ We can find that $y$ has roots at $x=0,3\pm\sqrt{3}$, or approximately $x=0, 1.3,4.7$.

The two graphs intersect at $(0,0)$ and touch at $(2,8)$.

Differentiating gives $\frac{dy}{dx} = -6x^2 + 24x - 12,$ so stationary points occur when \begin{align*} & -6x^2 + 24x - 12 = 0 \\ \iff & x^2 - 4x + 2 = 0 \\ \iff & x = 2 \pm \sqrt{2}. \end{align*}

So there are stationary points at $x \approx 3.4$ and $x \approx 0.6$.

So when $a=-2$, the graphs have the following shape.