Two classical hyperbolas in the Cartesian plane
The seemingly simple hyperbola 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 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 , pleasantly without calculus
The hyperbola
Here is the graph of the deceptively simple relation , or equivalently , which connects a number to its reciprocal . A typical point on the curve is , such as , and .
Note that the equation expresses the two asymptotes and .
The hyperbola
Here is another hyperbola: with equation . This does not define a function in the usual sense, because it cannot be expressed as a function . Instead this is an example of a more general relation between and .
Note that the equation expresses the two asymptotes and because .
Q1 (C): Can you see any similarities between the hyperbolas and ? Is there any way to get one from another?
Finding secants to
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 will help us to find a general equation for a tangent line to in the next step.
Suppose that and are two points on the hyperbola. To find the equation of the secant line we can first find the slope
and then write the equation as
for some unknown number .
When we substitute the point into this equation, we get
so that
It follows that the equation of the secant line is
This can be rewritten also in the form
Q2 (E): What are the and intercepts of the secant line?
Tangents to the hyperbola
A tangent line intersects a curve at only one point. So, to find the tangent line to the hyperbola at the point we can just set in the equation of the secant line. This gives
or, equivalently
Hidden in this seemingly innocent derivation is a key formula of calculus: that the slope of the tangent at the point is , which is the derivative of at the point .
Q3 (E): What are the and intercepts of the tangent line?
Q4 (M): Show that the product of the and intercepts is constant, independent of the point .
Q5 (M): Show that if is the point and then . This verifies another of Apollonius’ claims.
Answers
A1. Please let us know in the comments what you think.
A2. Setting we find the intercept is Setting we find the intercept is
A3. From the equation of the tangent line we find that the intercept is and the intercept is .
A4. From the previous example we have that the intercepts are and , and so their product is , which is constant.
A5. The quadrance between the points and is
The quadrance between the points and is
So
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