Tangents and normals to a rectangular hyperbola

The gradient

dy/dx = -c² x^-2 = -c²/x². At (ct, c/t): dy/dx = -c²/(c^2t2) = -1/t². The gradient is always negative, as the curve always slopes downwards.

Tangent and normal

Tangent: y - c/t = -(1/t²)(x - ct). Multiply by t²: x + t² y = 2ct.

Normal: gradient t², so y - c/t = t²(x - ct). Multiply by t: t³ x - ty = c(t⁴ - 1).

Typical follow-ups

• The tangent meets the axes at (2ct, 0) and (0, 2c/t), so the triangle it makes with the axes has area (1/2)(2ct)(2c/t) = 2c², whatever t is.
• The normal at one point usually meets the curve again: substitute the line into xy = c² and solve the quadratic, using the root you already know.

Worked example

The normal to xy = 16 at P(8, 2) meets the curve again at Q. Find Q.

1. c = 4 and t = 2. The normal gradient is t² = 4.
2. y - 2 = 4(x - 8), so y = 4x - 30.
3. x(4x - 30) = 16 gives 2x² - 15x - 8 = 0, (x - 8)(2x + 1) = 0.
4. x = 8 is P, so x = -1/2, and y = 4(-1/2) - 30 = -32.

Answer: Q(-1/2, -32)

Check yourself

1. Find the gradient of the tangent to xy = 16 at (2, 8).
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   -4. y = 16x^-1, dy/dx = -16/x² = -16/4 = -4.

2. The tangent to xy = 16 at (2, 8) meets the y-axis at (0, k). Find k.
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   16. y - 8 = -4(x - 2), so y = -4x + 16 and k = 16.

3. The tangent at any point of xy = 16 forms a triangle with the coordinate axes. Find its area.
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   32. The intercepts are 2ct and 2c/t, so the area is (1/2)(2ct)(2c/t) = 2c² = 32.

4. The normal to xy = 16 at (8, 2) meets the curve again at Q. Find the x-coordinate of Q. Give your answer as a fraction in its simplest form.
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   -1/2. The normal is y = 4x - 30. x(4x - 30) = 16 gives 2x² - 15x - 8 = (x - 8)(2x + 1) = 0, so x = -1/2.

5. Which is the equation of the tangent to xy = c² at (ct, c/t)?
   • x + t² y = 2ct
   • t² x + y = 2ct
   • x - t² y = 0
   • t³ x - ty = c(t⁴ - 1)
   Show answer

   x + t² y = 2ct. y - c/t = -(1/t²)(x - ct); multiply by t²: t² y - ct = -x + ct.

6. Find the gradient of the normal to xy = 9 at the point with parameter t = 3.
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   9. The tangent gradient is -1/t² = -1/9, so the normal gradient is t² = 9.

7. The normal to xy = 25 at (5, 5) meets the curve again at (p, p). Find p.
   Show answer

   -5. At t = 1 the normal gradient is 1, so the normal is y = x. It meets xy = 25 again at (-5, -5).

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Sources

• Pearson Edexcel International Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (issue 3), Unit FP1
• Pearson Edexcel IAL Further Pure Mathematics 1 scheme of work

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