Can we find the angle between these two tangents?

Question

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 $c$ .

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 $P$ .

Suggestion

Every parabola has a special point, called the focus. If you imagine turning the parabola into a mirror, and turning it to face the sun, then all of the sun’s rays would be reflected by the parabola through its focus. For the parabola, $y^2 = 4ax$ , the focus is at $(a,0).$

You might like to take a look at our Parabolic Mirrors resource.

What’s the equation of a line with gradient $m$ that passes through $(a,0)$ ? Where does this line intersect $xy=c^2$ ?

If the line is a tangent, how many times does it intersect the curve?

What does this tell us about $m$ ? Once we’ve found $m$ , can we find $P$ ?

If $P$ also lies on the parabola, prove that $a^4=2c^4$ …

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The blue curve here is the hyperbola $xy = c^2$ , while the red curve is the parabola $y^2=4ax$ .

The green line is the tangent to the hyperbola though the parabola’s focus $F$ , which touches the hyperbola at $P$ .

The values of $a^4$ and $2c^4$ are given. Can we use this to verify what the question asks us to show?

… and calculate the acute angle between the tangents to the two curves at $P$ .

How is the gradient of a line linked to the angle it makes with the $x$ -axis?

Or as an alternative method… the dot product of two vectors can help us find the angle between them.

Solution

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

$(a,0)$

y=m(x-a).

This will intersect the curve

$xy=c^2$

when:

\begin{align} x\times m(x-a) &= c^2 \not ag \\ \Longrightarrow \quad mx^2 -amx-c^2 &= 0. \end{align}

$y=m(x-a)$

is to be tangent to the hyperbola

$xy=c^2,$

then

$(1)$

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 \not ag \\ \Longrightarrow \quad m(a^2m+4c^2) &=0 \not ag \\ \Longrightarrow \quad m=-\frac{4c^2}{a^2}. \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 $P=\left(\frac{a}{2},\frac{2c^2}{a}\right).$

$P$

also lies on the parabola then

$P$

satisfies

$y^2=4ax,$

\begin{align} \left(\frac{2c^2}{a}\right)^2 &= 4a\left(\frac{a}{2}\right) \not ag \\ \Longrightarrow \quad \frac{4c^4}{a^2} &= 2a^2 \not ag \\ \Longrightarrow \quad 2c^4 &= a^4. \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 $\alpha - \theta$ .

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, $(grad-of-tangent)$ and $(c4a4)$ , 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.