The hyperbola

Cartesian equation

x²/a² - y²/b² = 1.

It crosses the x-axis at the vertices (±a, 0) and never crosses the y-axis (x = 0 gives y² = -b²).

Its asymptotes are y = ±(b/a)x: for large x and y the 1 is negligible, so x²/a² ≈ y²/b². When a = b it is a rectangular hyperbola with perpendicular asymptotes y = ±x (the F1 curve xy = c² is this shape rotated through 45°).

Parametric forms

• x = a sec t, y = b tan t, using sec² t - tan² t = 1. This gives both branches.
• x = a cosh t, y = b sinh t, using cosh² t - sinh² t = 1. Since cosh t ≥ 1, this gives only the right-hand branch (x ≥ a). Use x = -a cosh t for the left branch.

The hyperbolic functions get their name from exactly this link.

Standard form

Divide so the right-hand side is 1: 9x² - 4y² = 36 becomes x²/4 - y²/9 = 1, so a = 2 and b = 3, with asymptotes y = ±(3/2)x.

Worked example

Find the vertices and asymptotes of 16x² - 9y² = 144.

1. Divide by 144: x²/9 - y²/16 = 1
2. a = 3, b = 4

Answer: Vertices (±3, 0); asymptotes y = ±(4/3)x

Check yourself

1. What are the asymptotes of x²/16 - y²/9 = 1?
   • y = ±(3/4)x
   • y = ±(4/3)x
   • y = ±(9/16)x
   • x = ±4
   Show answer

   y = ±(3/4)x. a = 4 and b = 3, so y = ±(b/a)x = ±(3/4)x.

2. Which part of x²/a² - y²/b² = 1 is traced by x = a cosh t, y = b sinh t?
   • The left-hand branch only
   • The whole curve
   • The right-hand branch only
   • The asymptotes
   Show answer

   The right-hand branch only. cosh t ≥ 1, so x = a cosh t ≥ a: only points with x ≥ a are reached.

3. Find the x-coordinate of the point with t = π/4 on x = 2 sec t, y = 5 tan t, to 3 decimal places.
   Show answer

   2.828. x = 2/cos(π/4) = 2sqrt(2) = 2.828.

4. Find the gradient of the asymptote with positive gradient for 9x² - 4y² = 36.
   Show answer

   1.5. Standard form x²/4 - y²/9 = 1, so a = 2, b = 3 and the gradient is b/a = 1.5.

5. The point (5, k), where k > 0, lies on x²/16 - y²/9 = 1. Find k.
   Show answer

   2.25. 25/16 - k²/9 = 1, so k² = 9 × 9/16 = 81/16 and k = 9/4 = 2.25.

6. Why does x²/a² - y²/b² = 1 not cross the y-axis?
   • The asymptotes block it
   • It is symmetric in the y-axis
   • It only exists for y > 0
   • x = 0 gives y² = -b², which has no real solution
   Show answer

   x = 0 gives y² = -b², which has no real solution. Substituting x = 0 gives -y²/b² = 1.

7. Which is a rectangular hyperbola?
   • x²/4 - y²/9 = 1
   • x² - y² = 4
   • x²/4 + y²/4 = 1
   • x²/9 - y² = 1
   Show answer

   x² - y² = 4. x²/4 - y²/4 = 1 has a = b, so the asymptotes y = ±x are perpendicular.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (issue 3), Unit FP3

Last updated 8 November 2026. Spotted a mistake? Tell us.