The image in the warm-up shows one of a family of functions represented by the implicit equation \((x^2+2ay-a^2)^2=y^2(a^2-x^2)\).

Try changing \(a\) using the slider in the applet below.

What do you notice about other functions in the family?

A graph of an implicit curve

The initial image shown is the case where \(a=3\) (and is the same as the warm-up image).

Moving the slider to show the graph for other positive values of \(a\) we might notice that the shape of the graph seems to remain the same, but it does get bigger / smaller in line with the value of \(a\).

If we take the slider further to the left we notice that for negative values of \(a\) the shape of the graph seems to behave in the same way as for the corresponding positive values of \(a\), except that it is reflected in the \(x\)-axis.

Thinking more about the features for a particular value of \(a\), it looks as though the graph is symmetrical about the \(y\)-axis and that it has a maximum at \(y=a\). It looks like the graph is showing a closed loop or two curves that meet. It also seems that something interesting is happening at \(x=\pm a\).

Can you use a combination of the information provided by the graphical and algebraic representations to answer and explain some of the questions below?

  • What is special about the points:
    • \(x=a\)
    • \(x=-a\)
    • \(x=0\)

The first case, \(x=a\) appears to be a point where the graph touches the \(x\)-axis. It is interesting as it seems to be a ‘sharp’ point where two curves meet, rather than it being a part of a smooth, continuous curve.

Remember, we are making assumptions based on the graphical representation. It could be the case that if we zoomed in the graph looks like a smooth, continuous curve. Similarly, it could be that the graph is showing two distinct curves that come very close, but do not actually meet at all.

We will need to consider information that the algebraic representation can give us in order to get a better picture of how the graph is behaving.

We might like to confirm that \((a,0)\) is indeed a point on the graph (so far we have made an assumption based on the family of graphs we can see in the GeoGebra applet). If we substitute \(x=a\) into the implicit equation we have \[(a^2+2ay-a^2)^2=y^2(a^2-a^2),\] which can be simplified to \[4a^2y^2=0,\] which tells us that \(y=0\).

This tells us that \((a,0)\) is a point on the graph represented by \((x^2+2ay-a^2)^2=y^2(a^2-x^2)\). It does not tell us whether this is a point where the curves intersect the \(x\)-axis or just touch it.

How can we try to convince ourselves that the curves do not cross the \(x\)-axis?

The second case, \(x=-a\) looks as though it behaves in the same way as \(x=a\). We may notice that because all of the \(x\) terms in the implicit equation are of even power, the family of graphs will be symmetric about the \(y\)-axis.

Finally we consider the case \(x=0\). This is an interesting case as from the graphical representation we are expecting there to be two corresponding values of \(y\).

What do you predict the two values of \(y\) to be for \(x=0\)?

Substituting \(x=0\) into the implicit equation we have \[(2ay-a^2)^2=y^2a^2.\] Expanding the left-hand side and simplifying we find a quadratic equation for \(y\), \[3y^2-4ay+a^2=0.\]

We are expecting to have two solutions so finding a quadratic equation for \(y\) must be a good sign.

Does a quadratic equation always have two solutions?

Solving the quadratic equation we find that when \(x=0\), \(y=a\) or \(y=\frac{a}{3}\).

Is this what you expected? Does it match with your findings from the graphical representation?

  • Will each member of the family be represented by two curves? Why?

We have begun to address this question by thinking above about what happens when \(x=0\). The graphical representation suggests that all members of the family will be represented by two curves and the algebraic representation tells us that \(x=0\) corresponds to two values of \(y\).

We shouldn’t forget that \(a=0\) is also a member of this family. What happens when \(a=0\)?

What we haven’t yet shown is that there are two values of \(y\) for every \(x\).

We might try to do this algebraically by trying to take a similar approach as for the case when \(x=0\), attempting to rearrange the implicit equation in terms of \(y\). We might similarly expect to find a quadratic equation in \(y\).

We begin with the implicit equation \((x^2+2ay-a^2)^2=y^2(a^2-x^2)\). We are looking for an equation in terms of \(y\) but this is not a simple case of re-arranging because the \(y\) terms are ‘tangled’ up in the squared expression on the left-hand side.

We might try to simplify things by taking the square root of both sides of the equation, giving \[x^2+2ay-a^2=\pm y\sqrt{a^2-x^2}.\]

We have remembered to consider the positive and negative square roots. Why is it OK that the \(\pm\) only appears on one side of the equation?

Rearranging to collect \(y\) terms we have \[y\left(2a \pm \sqrt{a^2-x^2}\right) = a^2 - x^2,\] and then \[y=\frac{a^2-x^2}{2a \pm \sqrt{a^2-x^2}}.\]

Can you convince yourself that this form of the equation is equivalent to that presented in the problem? \[y=\frac{(x^2-a^2)(-2a \pm \sqrt{(a^2-x^2)})}{3a^2+x^2}\]

We begin with the implicit equation \((x^2+2ay-a^2)^2=y^2(a^2-x^2)\). In the case where \(x=0\) we then chose to expand the left-hand side.

Let’s try a similar approach which gives us \[x^4 +a^4 + 4ax^2y+4a^2y^2-4a^3y-2a^3x^2 = y^2(a^2-x^2).\]

If we now rearrange the left-hand side we have, \[4a^2y^2+4a(x^2-a^2)y-2a^2x^2+x^4+a^4=y^2(a^2-x^2),\] and collecting the \(y^2\) terms to reveal a quadratic equation in \(y\) we have \[(3a^2-x^2)y^2+4a(x^2-a^2)y-2a^2x^2+x^4+a^4=0.\]

Take a look at the last part of the quadratic equation, shown in blue below.

\[(3a^2-x^2)y^2+4a(x^2-a^2)y \color{blue}{-2a^2x^2+x^4+a^4}=0\]

We might notice that this could also be written as \((x^2-a^2)^2\). This might be useful as the expression \((x^2-a^2)\) appears elsewhere in the equation.

We now have, \[(3a^2-x^2)y^2+4a(x^2-a^2)y \color{blue}{+(x^2-a^2)^2}=0\]

Using the quadratic formula we could now re-write the equation in the form \(y=f(x)\). There will be a lot going on so it might make sense to consider simplifying the discriminant first.

Simplifying the equations we find \[y=\frac{(x^2-a^2)(-2a \pm \sqrt{(a^2-x^2)})}{3a^2+x^2}.\]

  • What is the domain and range of each member of the family?

From the graphical representation we expect the domain of each member of the family to be \(-a ≤ x ≤ a\). Looking at the equation in the form

\[y=\frac{(x^2-a^2)(-2a \pm \sqrt{a^2-x^2})}{3a^2+x^2}\] or \[y=\frac{a^2-x^2}{2a \pm \sqrt{a^2-x^2}},\] we might notice that because of the square root \(\sqrt{a^2-x^2}\) we require that \(x^2 ≤ a^2\) and can therefore confirm that \(-a ≤ x ≤ a\).

From the graphical representation again, we expect the range of each member of the family to be \(0 ≤ y ≤ a\).

Can we confirm this by referring back to the algebraic representations? How might we do this?

For example, what happens when \(x=0\)?