The point \(P\) on the hyperbola \(xy=c^2\) is such that the tangent to the hyperbola at \(P\) passes through the focus of the parabola \(y^2=4ax\). Find the coordinates of \(P\) in terms of \(a\) and \(c\).

Graph showing the parabola y squared = 4 a x, the hyperbola x y = c squared, and the tangent to the hyperbola at the point P where the parabola and hyperbola meet, which passes through (a,0), which is the focus of the parabola.
It is a standard result that the focus of the parabola \(y^2=4ax\) is at \((a,0)\). The general equation of a straight line with gradient \(m\) passing through \((a,0)\) is \[y=m(x-a).\] This will intersect the curve \(xy=c^2\) when: \[\begin{align} x\times m(x-a) &= c^2 \notag \\ \Longrightarrow \quad mx^2 -amx-c^2 &= 0. \label{eq:1} \end{align}\] If \(y=m(x-a)\) is to be tangent to the hyperbola \(xy=c^2,\) then \(\eqref{eq:1}\) will have a repeated root, which means the discriminant of this quadratic will be \(0\). This gives \[\begin{align} a^2m^2-4\times m \times (-c^2)&=0 \notag \\ \Longrightarrow \quad m(a^2m+4c^2) &=0 \notag \\ \Longrightarrow \quad m=-\frac{4c^2}{a^2}. \label{eq:grad-of-tangent} \end{align}\]

We have ignored the \(m=0\) solution, as this is the horizontal asymptote of \(xy=c^2\). We now have \[x=\frac{am\pm \sqrt{0}}{2m}=\frac{a}{2}.\] Since \(P\) lies on \(xy=c^2\), the corresponding \(y\)-coordinate is \[y=\frac{c^2}{x}=\frac{2c^2}{a}\] so \[P=\left(\frac{a}{2},\frac{2c^2}{a}\right).\]

If \(P\) also lies on the parabola, prove that \(a^4=2c^4\)

If \(P\) also lies on the parabola then \(P\) satisfies \(y^2=4ax,\) so \[\begin{align} \left(\frac{2c^2}{a}\right)^2 &= 4a\left(\frac{a}{2}\right) \notag \\ \Longrightarrow \quad \frac{4c^4}{a^2} &= 2a^2 \notag \\ \Longrightarrow \quad 2c^4 &= a^4. \label{eq:c4a4} \end{align}\]

… and calculate the acute angle between the tangents to the two curves at \(P\).

If a line of gradient \(m\) cuts the \(x\)-axis at an angle \(\theta\), then \(m = \tan \theta\). From the following diagram, we see that we need to find \(\alpha - \theta\).

The two curves again with tangents to both the parabola and the hyperbola at P drawn.

What is the gradient of the parabola, \(y^2 = 4ax\), at \(P\)? We can either differentiate implicitly or take the square root, \[y = 2\sqrt{ax} \implies \frac{dy}{dx} = 2\sqrt{a}\times\frac{1}{2}x^{-1/2} = \sqrt{\dfrac{a}{x}}.\]

Since the \(x\)-coordinate of \(P\) is \(\dfrac{a}{2}\), we have the gradient \(\tan \theta = \sqrt{\dfrac{a}{a/2}} = \sqrt{2}\).

From our earlier work, \(\eqref{eq:grad-of-tangent}\) and \(\eqref{eq:c4a4}\), we know \(\tan \alpha = -\dfrac{4c^2}{a^2} = -\dfrac{4}{\sqrt{2}} = -2\sqrt{2}\).

Now we can use a standard trig identity to find

\[\begin{align*} \tan(\alpha-\theta) &= \dfrac{\tan \alpha - \tan \theta}{1+\tan \alpha \tan \theta}\\ &= \dfrac{ -2\sqrt{2}-\sqrt{2}}{1-2\sqrt{2}\sqrt{2}} \\ &= \sqrt{2}.\\ \end{align*}\]

Hence the angle between the tangents at \(P\) is \(\arctan \sqrt{2}\approx 0.955^c \approx 54.7^\circ\).

It turns out that \(\theta=\alpha-\theta\) and the triangle in the above diagram is isosceles.

Instead of using the identity, we could have calculated \(\arctan{\sqrt{2}}\) and \(\arctan{\left(-2\sqrt{2}\right)}\) and taken the difference.

Alternatively, we could find the angle between the two lines using the dot product of the two direction vectors.

The tangent to the parabola has gradient \(\sqrt{2}\) so its direction vector can be written as \[\mathbf{a} = \begin{pmatrix}1 \\ \sqrt{2}\end{pmatrix}\] and the tangent to the hyperbola can be written as \[\mathbf{b} = \begin{pmatrix}1 \\ -2\sqrt{2}\end{pmatrix}.\]

The dot product can then be computed as \[\begin{align*} \mathbf{a}\cdot\mathbf{b} &= 1-2\sqrt{2}\sqrt{2}=-3 \\ &= |\mathbf{a}| \: |\mathbf{b}| \cos\theta = \sqrt{1+2}\sqrt{1+4\times2}\cos\theta \\ \implies\quad \cos\theta &= -\frac{1}{\sqrt{3}} \end{align*}\]

This actually gives us the obtuse angle but it’s then easy to find the acute one.