What is implied if these two parabolas touch exactly once?

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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 \(\pm\sqrt{k}\).

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

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

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

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

The two curves intersect when \[mx=k-(4-x^2)^2=k-(16-8x^2+x^4).\] Since \(m = 12\), this is equivalent to \[\begin{equation} x^4-8x^2+12x+(16-k)=0. \label{eq:1} \end{equation}\] 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.

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 \[1,\,-\frac{1}{2}\pm\frac{1}{2}\sqrt{13}.\]

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.

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).