Review question

# What is implied if these two parabolas touch exactly once? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5739

## Solution

It is given that the two curves $y=4-x^2 \qquad \text{and} \qquad mx=k-y^2,$ where $m>0$, touch exactly once.

1. In each of the following four cases, sketch the two curves on a single diagram, noting the coordinates of any intersections with the axes:

The first equation, $y = 4-x^2$ does not involve $k$ and $m$, so this curve does not vary.

We can rearrange the second curve, $mx=k-y^2$, into the form $x=-\dfrac{y^2}{m}+\dfrac{k}{m}$ and see that we have a quadratic curve in $y$, a ‘vertex-right’ parabola, with an $x$-intercept of $\dfrac{k}{m}$ and $y$-intercepts of $\pm\sqrt{k}$.

1. $k<0$;

The curves touch in the third quadrant, and do not intersect anywhere else.

1. $0< k <16$, $k/m<2$;

The curves touch in the first quadrant, and also intersect in the second and third quadrants.

1. $k>16$, $k/m>2$;

The curves touch in the first quadrant, and also intersect in the third and fourth quadrants.

1. $k>16$, $k/m<2$.

The curves touch in the first quadrant and also intersect in the first and third quadrants.

1. Now set $m = 12$. Show that the $x$-coordinate of any point at which the two curves meet satisfies $x^4-8x^2+12x+16-k=0.$
The two curves intersect when $mx=k-(4-x^2)^2=k-(16-8x^2+x^4).$ Since $m = 12$, this is equivalent to $\begin{equation} x^4-8x^2+12x+(16-k)=0. \label{eq:1} \end{equation}$

Let $a$ be a value of $x$ at the point where the curves touch. Show that $a$ satisfies $a^3-4a+3=0$ and hence find the three possible values of $a$.

If the curves touch at $x=a$, the gradients of the two curves are equal at this point. We have for the first curve $\frac{dy}{dx}=-2x$ and, differentiating implicitly for the second curve $12=-2y\frac{dy}{dx}.$ If they touch at $x = a$, $y = 4-a^2$, then $12 = -2y \times (-2a) = 4ay = 4a(4-a^2).$ This gives \begin{align*} 2a(4-a^2)-6&=0\\ \Longrightarrow \, a^3-4a+3&=0 \end{align*}

as required.

This can be found without calculus. Solving $y=4-x^2$ and $12x=k-y^2$ together gives $x^4-8x^2+12x+16-k=0.$

If we now equate this to $(x-a)^2(x-b)(x-c)=0$ (we have a double root at $a$), equate coefficients, and eliminate $b, c$ and $k$, we get $a^3-4a+3=0.$

By inspection, $a=1$ is a solution, so we take out $a-1$ as a factor and get $(a-1)(a^2+a-3)=0.$ The roots of the quadratic are $a=\frac{-1\pm\sqrt{13}}{2},$ so the possible values of $a$ are $1,\,-\frac{1}{2}\pm\frac{1}{2}\sqrt{13}.$

Derive also the equation $k=-4a^2+9a+16.$

Using the equation $\eqref{eq:1}$ above, and remembering that $a$ is a value of $x$, we obtain the equation $k = a^4 - 8a^2 + 12a + 16,$ and so \begin{align*} k &= 12a+16-8a^2+a^4\\ &= 16+12a-8a^2+a(4a-3)\\ &= 16+9a-4a^2, \end{align*}

where we’ve used $a^3=4a-3$ from the previous part to eliminate the $a^4$ term.

Which of the four sketches in part (i) arise?

When $a=1$, we have $k=16+9-4=21>16$ and $\dfrac{k}{m}=\dfrac{21}{12}<2$. This is therefore case (d).

Now $a=\dfrac{-1+\sqrt{13}}{2} \approx 1.30 \implies k \approx 20.9 > 16$, and $\dfrac{k}{m} \approx \dfrac{20.9}{12}<2$, and so we have situation (d).

On the other hand, $a=\dfrac{-1-\sqrt{13}}{2} \approx -2.30 \implies k \approx -25.9$, and so we have situation (a).

It’s worth noting that two of the values of $k$ we have here are very close to each other, $k = 21$, and $k \approx 20.9$.

Try zooming in on these two solutions using the GeoGebra applet below.