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 \(c\).
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 \(a^4=2c^4\)
… 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 \(\alpha - \theta\).
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.
