# Maltese Cross

Sketch the graph of $xy(x^2 - y^2) = x^2 + y^2$ consisting of four curves and a single point at the origin. Convert to polar form. Describe the symmetries of the graph.

The graph of points $(x,y)$ satisfying the equation

$$xy(x^2 - y^2) = x^2 + y^2$$

consists of four curves together with a single point at the origin.

You can use graphing software to sketch this graph but it is more of a challenge to see if you can sketch it for yourself and the steps in this question are designed to help you to do so.

(a) Prove that the polar equation of this graph is

$$r^4 \sin 4\theta = 4r^2.$$

(b) Deduce that there are 4 points on this graph at distance 2 from the origin and no points closer to the origin. Find the values of $\theta$ for which there are points on the graph and the values of $\theta$ for which there are no points on the graph.

(c) Substitute $y=px$ in the Cartesian equation and find an expression for $x^2$ in terms of $p$. Hence find the values of $p$ for which the lines $y=px$ do not cut the graph in points other than the origin. For other values of $p$, in how many points do the lines $y=px$ cut the graph?

(d) Prove that if the point $(a,b)$ lies on the graph then so do the points $(-a,-b)$, $(-b,a)$ and $(b,-a)$.

(e) What can you say about the symmetries of the graph.

You are told that the graph of points $(x,y)$ satisfying the equation

$$xy(x^2 - y^2) = x^2 + y^2$$

consists of four curves together with a single point at the origin.

To help you to get started, think of a point $P$ on the graph, at a distance $r$ from the origin, having coordinates $(x,y)$, where the line $OP$ joining $P$ to the origin makes an angle $\theta$ with the $x$-axis. Now, to find the polar equation, make the substitution

$$x = r \cos \theta,\ y = r \sin \theta$$

and don't be put off by the long expression you get. Use the trig identities for $\sin^2\theta $ and $\cos^2\theta$ and you should be able to simplify the expression to prove that the polar equation of this graph is

$$r^4 \sin 4\theta = 4r^2.$$

Clearly $r=0$ satisfies this equation. Is it possible to have $r< 2$?

What are the valuesof $\theta$ when $r=2$?

One more hint and all should be plain sailing: notice that $\sin 4\theta$ is never negative.

Jim sent us his solution:

I made the substitution $xy(x^2 - y^2) = x^2 + y^2$ that was given in the hint. This gave:

$$r^4\cos \theta \sin \theta (\cos^2\theta-\sin^2\theta)= r^2$$

I then subsituted the double angle formulae:

$$\cos2\theta=\cos^2\theta-\sin^2\theta$$ and $$\sin 2\theta=2\sin\theta\cos\theta$$

to get $r^4\sin4\theta=4r^2$

Next, I looked for points distance $2$ from the origin. So $r=2$. This meant that $sin 4\theta=1$, and sosince$\theta$ must be between $0$ and $360$ degrees,it has four solutions: $22.5$ degrees, $112.5$ degrees, $202.5$ degrees and $292.5$ degrees.

I rearranged theformula above to get $\sin4\theta=4/r^2$. Since there are only solutions to $\sin$ that are no more than $1$, $r$ must be at least $2$. This formula also implies that $\sin4\theta$ is never negative, and so $\theta$ can only only take the values in the ranges $0$ to $45$ degrees, $90$ to $135$ degrees, $180$ to $225$ degrees and $270$ to $315$ degrees.

When I made the substitution $y=px$, I got $x^2=\frac{1+p^2}{p(1-p^2)}$. In order for this line to only cut the graph at the origin, this must have no solutions. So we need $x^2< 0$. This meant that $1< p^2$, and so $|p|> 1$

*(Not quite right; see amendment below.)*

For the next part, I just substituted in the different co-ordinates, and found they satisfy the same equation as $(a,b)$ do, and so must be also on the graph.

Finally, I noticed that from the previous observation, that the graph has rotational symmetry about the origin,with an angle of $90$ degrees. I used this to sketch the graph, which I used a computer to confirm:

Image

Deaglan and Ciaran from St Paul's Catholic College agree with most of Jim's reasoning, but have pointed out an error in part C. You can read their amendment here.