What can we say if this hyperbola and parabola meet at right angles?

Question

The curves $y^2=4ax$ and $xy=c^2$ intersect at right angles. Prove that

$c^4=32a^4$ and

if the tangent and normal to either curve at the point of intersection meet the $x$ -axis at $T$ and $G$ , then $TG=6a$ .

Suggestion

The curves $y^2=4ax$ and $xy=c^2$ intersect at right angles. Prove that (i) $c^4=32a^4$

What do the curves $y^2=4ax$ and $xy=c^2$ look like? Could we sketch a diagram?

If the curves meet at $(p,q)$ , what are the gradients of the curves there?

What condition on the slopes of two lines corresponds to them being at right angles?

What are the equations of the tangent and normal?

Where do these lines meet the $x$ -axis?

Solution

The curves $y^2=4ax$ and $xy=c^2$ intersect at right angles. Prove that (i) $c^4=32a^4$

Let the point of intersection be

$(p,q)$

. Thus

\begin{equation} q^2 = 4ap \text{, and } pq=c^2. \end{equation}

Using implicit differentiation on $y^2 =4ax$ , we have $2y\dfrac{dy}{dx} = 4a$ , so the gradient at $(p,q)$ is $\dfrac{dy}{dx} = \dfrac{2a}{q}$ .

Similarly, differentiating $xy=c^2$ , we have $y+x\dfrac{dy}{dx} = 0$ , so $\dfrac{dy}{dx}= \dfrac{-y}{x}$ , and the gradient at $(p,q)$ is $\dfrac{-q}{p}$ .

Since the curves are at right angles at $(p,q)$ , we have $\dfrac{2a}{q}\times\dfrac{-q}{p}=-1$ , which simplifies to $p = 2a$ .

Combining this with $(intersection)$ we have $q^2 = 8a^2$ , and since $p^2q^2=c^4$ we have $(2a)^28a^2 = c^4$ , or $c^4=32a^4$ , as required.

Prove that (ii) if the tangent and normal to either curve at the point of intersection meet the $x$ -axis at $T$ and $G$ , then $TG=6a$ .

The tangent and normal both pass through $(p,q)$ and have gradients $\dfrac{2a}{q}$ and $\dfrac{-q}{p}$ . So we can write their equations as $x-p = \dfrac{2a}{q}(y-q),\qquad\text{and}\quad x-p = \dfrac{-q}{p}(y-q) .$

Let these lines meet the $x$ -axis at $(x_1,0)$ and $(x_2,0)$ . Then we have $x_1-p = -\dfrac{2a}{q}q,\qquad\text{and}\quad x_2-p = \dfrac{q}{p}q .$

Subtracting these two and substituting from $(intersection)$ , $x_2-x_1 = \frac{q^2}{p}+2a = \frac{4ap}{p}+2a = 6a$ as required.

We have the two lines $x-p = m_1(y-q),\qquad x-p = m_2(y-q)$ where $m_1$ is the gradient of the hyperbola, and $m_2$ is the gradient of the parabola, both evaluated at $(p,q)$ . These meet the $x$ -axis when $y=0$ . Let these points be at $x_1$ and $x_2$ respectively -- we can be sure from the graph that $x_1>x_2$ . Subtracting the two equations, we get $(x_1-p) - (x_2-p) = -q(m_1-m_2)$ or $x_1 - x_2 = -q(m_1-m_2).$ Now $m_1=-\dfrac{q}{p}$ and $m_2=\dfrac{q}{2a}$ so $x_1 - x_2 = -q\left(-\frac{2a}{q}-\frac{q}{2a}\right) = \frac{4a^2+q^2}{2a} = \frac{4a^2 + 8a^2}{2a} = 6a,$ as required.