We will derive the equation of the hyperbola from its definition. The hyperbola is the locus of all points whose difference of the distances to two foci is contant. The equation of the hyperbola is \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) or \(-\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) depending on the orientation. We will use the first equation in which the transverse axis is the x-axis.

We will assume we already know that this difference is equal to 2a. We could let it equal some constant d but that is the same as letting it equal 2a and assuming we do not yet know what the a represents. Figure 1 depicts the parts of a hyperbola.

Figure 1: The parts of a hyperbola (\(|c| > |a|\))

We will place our foci at (–c, 0) and (c, 0) so that the major axis of the hyperbola is the x-axis. We can do a similar exercise if the major axis is the y-axis.

From the definition, we have the following equation and the steps that follow:

(i) \(\sqrt{(x-c)^2+y^2} - \sqrt{(x+c)^2+y^2} = 2a\)

(ii) \(2a + \sqrt{(x+c)^2+y^2} = \sqrt{(x-c)^2+y^2}\)   (Rearrange a bit)

(iii) \(4a^2 + 4a\sqrt{(x+c)^2+y^2} + (x+c)^2 + \cancel{y^2} = (x-c)^2 + \cancel{y^2}\)   (Square both sides)

(iv) \(4a^2 + 4a\sqrt{(x+c)^2+y^2} + \cancel{x^2} + 2cx + \cancel{c^2} + \cancel{y^2} = \cancel{x^2} - 2cx + \cancel{c^2} + \cancel{y^2}\)

(v) \(4a^2 + 4a\sqrt{(x+c)^2+y^2} + 4cx = 0 \)   (Clean up a bit)

(vi) \(a + \frac{c}{a}x = -\sqrt{(x+c)^2+y^2} \)   (Rearrange to isolate the radical again and divide by a)

(vii) \(a^2 + \cancel{2cx} + \frac{c^2}{a^2}x^2 = x^2 + \cancel{2cx} + c^2 +y^2 \)   (Square both sides again)

(viii) \(a^4 + c^2x^2 = a^2x^2 + a^2c^2 + a^2y^2 \)   (Clean up and multiply both sides by a²)

(ix) \((c^2-a^2)x^2 - a^2y^2 = a^2(c^2-a^2) \)   (Collect like terms)

(x) \(b^2x^2 - a^2y^2 = a^2b^2 \)   (Let \(c^2 - a^2 = b^2\))

(xi) \( \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \)   (Divide both sides by \(a^2b^2\))

Equation (xi) represents the equation of an hyperbola. As of yet, we only know that 2a represents the constant sum of the hyperbola, but do not know that this is the length of the major axis. We also know \(a^2 + b^2 = c^2\), but we do not know that 2b represents the length of the minor axis. However, we can easily determine these.

Equation (ix) is the exact same equation we would derive for an ellipse. If we swap the radicals in Equation (ii) and square both sides, the equations and the steps that follow are the exact same as an ellipse.

The only way to differentiate a hyperbola from an ellipse is to recognize in step (ix) that c > a and let \(c^2 - a^2 = b^2\) instead of \(a^2 - c^2 = b^2\). Therefore, in step (i), the negative sign makes no difference. In Figure 1, you can see that the foci are located beyond the distance |a| because |c| is greater than |a|.

The Distances

The Value a

How do we know the constant difference of a hyperbola is 2a? We used the value 2a to derive the equation of the hyperbola. However, we can establish this again. First, let’s find the x-intercepts because those are easy to find. (Note that there are no y-intercepts for a hyperbola with a horizontal transverse axis.) We let y equal 0 in the hyperbola equation to find the x-intercepts.

(i) \( \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 \)

(ii) \( \frac{x^2}{a^2}+0=1 \)

(iii) \( x^2 = a^2 \)

(iv) \( x = \pm a \)

Therefore, the two x-intercepts are (–a, 0) and (a, 0). And these are two points on the hyperbola. Therefore, their differences of the distances from the foci should equal the constant difference.

The distance from the (–c, 0) to (a, 0) is a + c. This is the distance from one focus to a point on the hyperbola. The distance from (a, 0) to (c, 0) is ca. This is the distance from the same point on the hyperbola to the other focus. The difference between these two distances is \((a + c) - (c - a) = a + c - c + a = 2a \).

The Value b

Now, we know that 2a is the length of the horizontal axis and the constant difference of the hyperbola. In step (x), we let \(c^2 = a^2 + b^2\). The value b² ended up as the denominator of y² term. Unlike the ellipse, we cannot let x = 0 because the hyperbola has no values between (–a and a).

To find the significance of b, we draw a circular arc with the origin as the center and c as the radius.

Figure 2

In Figure 2, the circular arc, which has a length of c, crosses a perpendicular line drawn at (a, 0) at point (a, b). Remember we let \(a^2 + b^2 = c^2\) in step (x). Therefore, the intersection of the circular arc and the perpendicular arc is the point (a, b).

Still, what is the significant of some point (a, b) on the perpendicular? Although it may seem the point is random, the asymptotes of the hyperbola passes through this point because the slopes of the two asymptotes are \(\pm\dfrac{b}{a}\).

Figure 3

Figure 3 shows the 2 asymptotes of the hyperbola. The equations of the asymptotes are \(y = \pm\dfrac{b}{a}x\).