Where do two tangents cross, and where do two normals cross?

Where do two tangents cross, and where do two normals cross?

Question

The curve $C$ has equation $xy=\frac{1}{2}$. The tangents to the curve $C$ at the distinct points $P(p,\frac{1}{2p})$ and $Q(q,\frac{1}{2q})$, where $p$ and $q$ are positive, intersect at $T$ and the normals to $C$ at these points intersect at $N$. Show that $T$ is the point $\left(\frac{2pq}{p+q},\frac{1}{p+q}\right).$ In the case $pq=\frac{1}{2}$, find the coordinates of $N$. Show (in this case) that $T$ and $N$ lie on the line $y=x$ and are such that the product of their distances from the origin is constant.

Suggestion

Sketch a diagram to help visualise the information given.

What is the gradient of $xy=\frac{1}{2}$ at a point? How can we then find the equation of a tangent to the curve at a certain point?

Once we have both tangents, how can we find out where they intersect?

If we know the gradient of a line, what is the gradient of a normal to that line?

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The blue curve here is $xy = \dfrac{1}{2}$. You can vary the value of $p$ to move the points on the curve.

The tangents at $P$ and $Q$ are shown in red, while the normals at these points are in green.

The rectangle has side lengths proportional to $|OT|$ and $|ON|$.

Solution

The curve $C$ has equation $xy=\dfrac{1}{2}$. The tangents to the curve $C$ at the distinct points $P(p,\dfrac{1}{2p})$ and $Q(q,\dfrac{1}{2q})$, where $p$ and $q$ are positive, intersect at $T$ and the normals to $C$ at these points intersect at $N$. Show that $T$ is the point $\left(\dfrac{2pq}{p+q},\dfrac{1}{p+q}\right).$

We have the points $P(p,\dfrac{1}{2p})$ and $Q(q,\dfrac{1}{2q})$ where, without loss of generality, $0< p< q$ as in the diagram. We can rearrange the equation of the curve to $y = \dfrac{1}{2x}$ and then differentiate to obtain

$$\frac{dy}{dx} = -\frac{1}{2x^2}.$$

At the point $P$, with coordinates $(p, \dfrac{1}{2p})$, the gradient is therefore $-\dfrac{1}{2p^2}$. Thus the equation of the tangent through $P$ is $\left(y-\frac{1}{2p}\right)=-\frac{1}{2p^2}(x-p),$ that is, $y=\frac{1}{p}-\frac{1}{2p^2}x.$ Similarly, the equation of the tangent through $Q$ is $y=\dfrac{1}{q}-\dfrac{1}{2q^2}x.$

The two tangents intersect when

$$\frac{1}{p}-\frac{1}{2p^2}x=\frac{1}{q}-\frac{1}{2q^2}x.$$

i.e. when

$$x\left(\frac{1}{2q^2}-\frac{1}{2p^2}\right) = \frac{1}{q}-\frac{1}{p},$$

We can simplify the left-hand bracket, using the difference of two squares, to give

$$\frac{1}{2q^2}-\frac{1}{2p^2}=\frac{1}{2} \left(\left(\frac{1}{q}\right)^2 -\left(\frac{1}{p}\right)^2\right) = \frac{1}{2}\left(\frac{1}{q}-\frac{1}{p}\right)\left(\frac{1}{q} + \frac{1}{p}\right) = \frac{1}{2}\left(\frac{1}{q}-\frac{1}{p}\right)\left(\frac{p+q}{pq} \right).$$

Then the equation involving the

$x$

-coordinate of

$T$

simplifies to

$$\frac{1}{2} x \left(\frac{1}{q}-\frac{1}{p}\right)\left(\frac{p+q}{pq} \right) = \left(\frac{1}{q}-\frac{1}{p}\right)$$

and so, dividing through by the bracketed expression on both sides (which is not zero since

$p \neq q$

) and making

$x$

the subject, we see that

$$x = \frac{2pq}{p+q}.$$

The

$y$

-coordinate of

$T$

is then

$$\begin{align*} y&=\dfrac{1}{p}-\dfrac{1}{2p^2}\times\dfrac{2pq}{p+q}\\ &=\dfrac{1}{p}\left(1-\dfrac{q}{p+q}\right)=\dfrac{1}{p+q}, \end{align*}$$

as required.

In the case $pq=\dfrac{1}{2}$, find the coordinates of $N$.

It would be helpful to find the equations of the normals to the curve at $P$ and $Q$.

We know that (gradient of the normal) $\times$ (gradient of the tangent) at any point on the curve is $-1$.

We can therefore use what we know about the gradients of the tangents to the curve at $P$ and $Q$.

The gradients of the normals through $P$ and $Q$ are $2p^2$ and $2q^2$, respectively. The equation of the normal through $P$ is therefore $\left(y-\frac{1}{2p}\right)=2p^2(x-p),$ which we can rearrange to give $y=2p^2x+\frac{1}{2p}-2p^3.$ Similarly, the normal through $Q$ has equation $y = 2q^2x + \frac{1}{2q} - 2q^3.$

The two normals intersect at

$N$

, whose

$x$

coordinate satisfies

$$2p^2x+\frac{1}{2p}-2p^3=2q^2x+\frac{1}{2q}-2q^3,$$

that is,

$$\begin{align*} 2(p-q)(p+q)x &= \dfrac{1}{2}\left(\dfrac{1}{q}-\dfrac{1}{p}\right)+2(p^3-q^3)\\ &= \dfrac{1}{2}\left(\dfrac{p-q}{pq}\right)+2(p-q)(p^2+pq+q^2) \end{align*}$$

and so $x = \frac{1}{p+q}\left(\frac{1}{4pq}+p^2+pq+q^2\right). \qquad (p\neq q)$

Using

$pq=\dfrac{1}{2}$

, this becomes

$$\begin{align*} x&=\dfrac{1}{p+q}\left(\dfrac{(pq)^2}{pq}+p^2+pq+q^2\right)\\ &=\dfrac{1}{p+q}\left(p^2+2pq+q^2\right)\\ &=\dfrac{1}{p+q}(p+q)^2\\ &=p+q. \end{align*}$$

Substituting back into the equation of the normal through $P$ we find $y=2p^2(p+q)+\frac{1}{2p}-2p^3=2p^2q+\frac{1}{2p}.$ Again, using $pq=\dfrac{1}{2}$ we find $y=2p\left(\frac{1}{2}\right)+q=p+q.$ Thus $N$ has coordinates $(p+q,p+q)$ when $pq=\dfrac{1}{2}$.

Show (in this case) that $T$ and $N$ lie on the line $y=x$…

When $pq=\dfrac{1}{2}$, $T$ has coordinates $\left(\dfrac{1}{p+q},\dfrac{1}{p+q}\right)$, so lies on $y=x$. And we have just shown that $N$ has coordinates $(p+q,p+q)$, which also lies in the line $y=x$.

Alternatively, when $pq = \dfrac{1}{2}$, then $P =\left(p, \dfrac{1}{2p}\right)$ and $Q=\left(q, \dfrac{1}{2q}\right) = \left(\dfrac{1}{2p}, p\right)$, which means by symmetry that $T$ and $N$ lie on $y = x$.

…and are such that the product of their distances from the origin is constant.

We need a way to find the distance of a point from the origin. We can use Pythagoras’s theorem to do this.

The point with coordinates $(a,b)$ is at distance $\sqrt{a^2 + b^2}$ from the origin.

The distance of $T$ from the origin is $\vert OT \vert = \sqrt{\frac{2}{(p+q)^2}}=\frac{\sqrt{2}}{p+q}.$ Similarly, $\vert ON \vert = \sqrt{2(p+q)^2}=\sqrt{2}(p+q).$ Thus the product of their distances from the origin is $\vert OT \vert \times \vert ON \vert = \frac{\sqrt{2}}{p+q} \cdot \sqrt{2}(p+q)=2,$ which is a constant.