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1.6
A graph of y=1/x and x^2-y^2=1
What do these 2 hyperbolas have in common?

Two classical hyperbolas in the Cartesian plane

The seemingly simple hyperbola \(\normalsize{y=1/x}\) has many interesting aspects. It is our basic model for an inverse relation, and it embodies the fundamental arithmetical operation of taking the reciprocal of a number. It has a sister hyperbola with equation \(\normalsize{x^2-y^2=1}\) that features prominently in Einstein’s special theory of relativity.

In this step we

  • look at the geometry of these two curves

  • find equations for secants and tangents for \(\normalsize{y=1/x}\), pleasantly without calculus

The hyperbola \(\normalsize{y=1/x}\)

Here is the graph of the deceptively simple relation \(\normalsize{y=1/x}\), or equivalently \(\normalsize{xy=1}\), which connects a number \(\normalsize{x}\) to its reciprocal \(\normalsize{1/x}\). A typical point on the curve is \(\normalsize{A=[t,1/t]}\), such as \(\normalsize [1,1]\), \(\normalsize [2,1/2]\) and \(\normalsize [3,1/3]\).

Graph of y=1/x

Note that the equation \(\normalsize{xy=0}\) expresses the two asymptotes \(\normalsize{y=0}\) and \(\normalsize{x=0}\).

The hyperbola \(\normalsize{x^2-y^2=1}\)

Here is another hyperbola: with equation \(\normalsize{x^2-y^2=1}\). This does not define a function in the usual sense, because it cannot be expressed as a function \(\normalsize{y=f(x)}\). Instead this is an example of a more general relation between \(\normalsize{x}\) and \(\normalsize{y}\).

Graph of x^2-y^2=1 with asymptotes

Note that the equation \(\normalsize{x^2 - y^2 = 0}\) expresses the two asymptotes \(\normalsize{y=x}\) and \(\normalsize{y=-x}\) because \(\normalsize{x^2 - y^2 = (x-y)(x+y)}\).

Q1 (C): Can you see any similarities between the hyperbolas \(\normalsize{y=1/x}\) and \(\normalsize{x^2-y^2=1}\)? Is there any way to get one from another?

Finding secants to \(\normalsize{y=1/x}\)

You’ll recall that any line that intersects a curve in 2 places is called a secant line. Finding the general equation for a secant line through our hyperbola \(\normalsize{y=1/x}\) will help us to find a general equation for a tangent line to \(\normalsize{y=1/x}\) in the next step.

Suppose that \(\normalsize{A=[t,1/t]}\) and \(\normalsize{B=[u,1/u]}\) are two points on the hyperbola. To find the equation of the secant line \(\normalsize{AB}\) we can first find the slope

\[\Large{\frac{1/u-1/t}{u-t}=-\frac{1}{ut}}\]

and then write the equation as

\[\Large{y=-\frac{1}{ut}x + b}\]

for some unknown number \(\normalsize{b}\).

When we substitute the point \(\normalsize{A=[t,1/t]}\) into this equation, we get

\[\Large{\frac{1}{t}=-\frac{1}{u} + b}\]

so that

\[\Large{b= \frac{1}{t} + \frac{1}{u} }.\]

It follows that the equation of the secant line is

\[\Large{y=-\frac{1}{ut}x + \frac{1}{t} + \frac{1}{u} }.\]

This can be rewritten also in the form

\[\Large{x+tuy=t+u}.\]

Q2 (E): What are the \(\normalsize{x}\) and \(\normalsize{y}\)-intercepts of the secant line?

Tangents to the hyperbola \(\normalsize{y=1/x}\)

A tangent line intersects a curve at only one point. So, to find the tangent line to the hyperbola at the point \(\normalsize{A=[t,1/t]}\) we can just set \(\normalsize{t=u}\) in the equation of the secant line. This gives

\[\Large{y=-\frac{1}{t^2}x + \frac{2}{t} }\]

or, equivalently

\[\Large{x+t^2y=2t}.\]

Hidden in this seemingly innocent derivation is a key formula of calculus: that the slope of the tangent at the point \(\normalsize A=[t,1/t]\) is \(\normalsize - \frac{1}{t^2}\), which is the derivative of \(\normalsize y=\frac{1}{x}\) at the point \(\normalsize x=t\).

Pic of y=1/x with point X on it, tangent line shown and the points P and R

Q3 (E): What are the \(\normalsize{x}\) and \(\normalsize{y}\)-intercepts of the tangent line?

Q4 (M): Show that the product of the \(\normalsize{x}\) and \(\normalsize{y}\)-intercepts of the tangent line is constant, independent of the point \(\normalsize{A=[t,1/t]}\).

Q5 (M): Show that if \(\normalsize{P}\) is the point \(\normalsize{[2t,0]}\) and \(\normalsize{R=[0,2/t]}\) then \(\normalsize{Q( A,P)=Q(A,R)}\), where \(A\) is the point on the hyperbola. This verifies another of Apollonius’ claims.

Answers

A1. Please let us know in the comments what you think.

A2. Setting \(\normalsize{x=0}\) we find the \(\normalsize{y}\)-intercept is \(\normalsize{\frac{t+u}{tu}}.\) Setting \(\normalsize{y=0}\) we find the \(\normalsize{x}\)-intercept is \(\normalsize{t+u}.\)

A3. From the equation of the tangent line \(\normalsize{y=-\frac{1}{t^2}x + \frac{2}{t} }\) we find that the \(\normalsize{y}\)-intercept is \(\normalsize{2/t}\) and the \(\normalsize{x}\)-intercept is \(\normalsize{2t}\).

A4. From the previous example we have that the intercepts are \(\normalsize{2/t}\) and \(\normalsize{2t}\), and so their product is \(\normalsize{2/t \times 2t= 4}\), which is constant.

A5. The quadrance between the points \(\normalsize A=[t,1/t]\) and \(\normalsize P=[2t,0]\) is

\[\Large Q(A,P)=(2t-t)^2+(0-\frac{1}{t})^2=t^2+\frac{1}{t^2}.\]

The quadrance between the points \(\normalsize A\) and \(\normalsize R=[0,2/t]\) is

\[\Large Q(A,R)= (0-t)^2+(\frac{2}{t}-\frac{1}{t})^2 = t^2+\frac{1}{t^2}.\]

So \(\normalsize Q( A,P)=Q( A,R).\)

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This article is from the free online course:

Maths for Humans: Inverse Relations and Power Laws

UNSW Sydney