Can someone please help with this question?

P(2p,2/p) and Q(2q, 2/q) are two points on the rectangular hyperbola xy=4. M is the midpoint of PQ.

Find the equation of the locus of M if PQ is a tangent to the parabola y^{2}=4x

Essentially, I determined the locus to be y^{2}=-x/4 (not sure if this is right as I don't have solutions). However, I'm conscious of the fact that there may be restrictions, but I can't seem to find what these may be. How do I determine the restrictions, if any?

Thank you.

That locus is correct, I think i did that question in a cssa paper. Does the question ask to specify restrictions? I don't see what the restrictions would be as you've eliminated the parameters

My working out and

GeoGebra are claiming \(y^2 = -\frac{x}{2}\). Note that the line(s) demonstrating that \(PQ\) is a tangent to \(y^2=4x\) is hidden, but you can put it back in there.

\[m_{PQ} = \frac{\frac2q-\frac2p}{2q-2p}=-\frac1{pq}\\ \text{so }PQ\text{ has equation }x+pqy=2(p+q).\]

\[ \text{For this to be tangential to }y^2=4x\text{, upon substitution we'd require}\\ \text{the point of intersection to satisfy}\\ \frac{y^2}{4}+pqy-2(p+q)=0. \]

\[ \text{The tangent will only have one point of intersection with the parabola}\\ \text{and hence this quadratic in }y\text{ has only one solution.}\\ \text{Setting the discriminant equal to 0 gives the constraint}\\ \boxed{p^2q^2+2(p+q) = 0}. \]

\[ \text{The coordinates of }M\text{ satisfy}\\ \begin{align*}x&= \frac{2p+2q}{2}=p+q\\ y&= \frac{\frac2p+\frac2q}2 = \frac{q+p}{pq}\end{align*} \]

\[ \text{Hence to find the locus of }M\text{, we first see that}\\ y=\frac{x}{pq} \implies pq = \frac{x}{y}. \]

\[ \text{Subbing this and }x=p+q\text{ into the constraint found above gives}\\ \frac{x^2}{y^2}+2x=0 \implies\boxed{ y^2 = -\frac{x}{2}} \]

Restrictions on this locus are not so trivially found. First note that the relationship between \(p\) and \(q\), when treated as a quadratic in \(q\), gives \( q=\frac{-1\pm \sqrt{1-2p^3}}{p^2}\). The expression under the square root shows that \(p < 2^{-1/3}\).

Close examination of the GeoGebra output shows that the restrictions on the locus appear to differ, depending on if the positive or negative root was taken as the correct one. In general, restrictions

**do** apply as the locus will not include

**every** point on its defining equation.

Here, taking the positive root appears to force the condition that \(y > 0\), i.e. we only consider the upper branch of the parabola \(y^2=-\frac{x}{2}\).

Taking the negative root gives a condition very difficult to describe. There are subcases to consider. When \(p < 0\), it can be shown that the locus is restricted to \(y <- 2^{-1/3}\). The significance of \(-2^{-1/3}\) is that it is the point of intersection between the locus \(y^2=-\frac{x}{2}\) and the original hyperbola \(xy=4\). On the other hand when \(0 < p < 2^{1/3}\), it instead appears as though the locus will always be on the upper branch of the parabola, with a new condition that \(x < -2^{-1/3} \) instead.

So overall (combining the two), the locus would be \(y^2=-\frac{x}{2}\), subject to the restriction that the parabolic arc from \( \left(-2^{5/3}, -2^{1/3}\right) \) to the origin \((0,0)\) is excluded. The exclusion includes the origin, but not the other endpoint.

The main thing making this so awkward is that

**sometimes** \(P\) and \(Q\) will be in the same quadrant, and

**other times** they won't be. Furthermore, each value of \(p\) corresponds to

**two** values of \(q\), and it appears as though the algebra works out differently depending on which is taken. For that reason, if you were asked to find the restrictions in an exam, you would be given hints. In general, excluded points on the locus do not need to be found unless the examination has specifically requested for it.