Quadratic Equations II: Tangents & Normal to Curves – SS2 Further Mathematics Lesson Note (First Term)

A tangent to a curve is the straight line that touches the curve at a point and has the same gradient as the curve at that point.

Behavioural objectives

By the end of the lesson, students should be able to:

1. state that dy/dx at a point on a curve gives the gradient of the tangent at that point;
2. find the equation of the tangent to a curve at a given point;
3. define the normal to a curve and state the relationship m₁ = -1/m;
4. find the equation of the normal to a curve at a given point;
5. find the equation of a tangent to a curve given implicitly, such as x²y + xy³ + 3x – 13 = 0.

What is the gradient of a tangent to a curve?

The gradient of the tangent at a point is the value of dy/dx at that point.

• Gradient function: dy/dx is the gradient function of the curve.
• Value at a point: put x = x₁ into dy/dx to get the gradient of the tangent at x₁.
• Notation: this value is written as dy/dx at x = x₁, and we call it m.
• Equation of a line: the line of gradient m through (x₁, y₁) is y – y₁ = m(x – x₁).
• Tangent: use this same formula with m = dy/dx at the point of contact.

What is the normal to a curve?

The normal to a curve is the straight line perpendicular to the tangent at the point of contact.

• Tangent gradient: m is the gradient of the tangent at the point of contact.
• Normal gradient: m₁ is the gradient of the normal, and m₁ = -1/m.
• Reason: the gradients of two perpendicular lines multiply to give -1, so m × m₁ = -1.
• Equation of the normal: at (x₁, y₁), it is y – y₁ = -1/m (x – x₁).

How do the tangent and the normal compare?

What happens when the tangent gradient is zero?

• Horizontal tangent: if m = 0, the tangent is the line y = y₁.
• Vertical normal: -1/m is undefined, so the normal is the vertical line x = x₁.

How do you find the equation of a tangent and a normal?

Follow these steps in order.

1. Differentiate the equation of the curve to get dy/dx.
2. Find y when x is given (or x when y is given) to get the point (x₁, y₁).
3. Substitute x₁ into dy/dx to get m.
4. Write the tangent: y – y₁ = m(x – x₁).
5. For the normal, change the gradient to -1/m and write y – y₁ = -1/m (x – x₁).
6. Simplify each equation.

How do you find the tangent and normal to a curve? Worked examples

Example 1: tangent and normal at x = 1

Find the equation of the tangent and the normal to the curve y = 2x³ – x² + 3x + 1 at the point x = 1.

1. Differentiate: dy/dx = 6x² – 2x + 3.
2. At x = 1, dy/dx = 6 – 2 + 3 = 7, so the tangent gradient is m = 7.
3. At x = 1, y = 2 – 1 + 3 + 1 = 5, so the point is (1, 5).
4. Tangent: y – 5 = 7(x – 1).
5. Expand: y – 5 = 7x – 7.
6. The tangent is y = 7x – 2.
7. Normal gradient: m₁ = -1/7.
8. Normal: y – 5 = -1/7 (x – 1).
9. Multiply by 7: 7(y – 5) = -(x – 1), so 7y – 35 = -x + 1.
10. The normal is 7y + x – 36 = 0.

How do you find a tangent when the curve is given implicitly?

An implicit curve has x and y mixed in one equation, so differentiate each term with respect to x.

• Terms in x only: differentiate as usual, so 3x gives 3.
• Terms in y: differentiate and multiply by dy/dx, so y³ gives 3y²(dy/dx).
• Products: use the product rule, so x²y gives 2xy + x²(dy/dx).

Example 2: tangent to a curve given implicitly

Find the equation of the tangent to the curve x²y + xy³ + 3x – 13 = 0 at the point (1, 2).

1. Check the point: 1(2) + 1(8) + 3 – 13 = 0, so (1, 2) lies on the curve.
2. Differentiate with respect to x: 2xy + x²(dy/dx) + 3xy²(dy/dx) + y³ + 3 = 0.
3. Collect the dy/dx terms: (x² + 3xy²)(dy/dx) = -2xy – y³ – 3.
4. Make dy/dx the subject: dy/dx = (-2xy – y³ – 3)/(x² + 3xy²).
5. Put x = 1 and y = 2: m = (-4 – 8 – 3)/(1 + 12) = -15/13.
6. Tangent: y – 2 = -15/13 (x – 1).
7. Multiply by 13: 13(y – 2) = -15(x – 1), so 13y – 26 = -15x + 15.
8. The tangent is 13y + 15x – 41 = 0.

What mistakes should students avoid?

• Using the wrong point: find y from the curve’s equation, not from the tangent.
• Forgetting the negative sign: the normal gradient is -1/m, not 1/m.
• Missing dy/dx on y terms: in implicit curves, every y term needs the factor dy/dx.
• Skipping the check: put the point into your final equation to confirm it fits.

Evaluation

1. What does the value of dy/dx at a point on a curve give?
2. Define the normal to a curve.
3. If the gradient of the tangent at a point is m, write down the gradient of the normal and the equation of the normal at (x₁, y₁).
4. Find:
   1. the equation of the tangent to the curve y = x³ + x² – 8x + 2 at the point A(1, -4);
   2. the coordinates of the point where the tangent meets the x-axis.

Assignment

1. A curve is defined by f(x) = x³ – 6x² – 15x – 1. Find:
   1. the derivative f'(x) with respect to x;
   2. the gradient of the curve at the point where x = 1;
   3. the equation of the tangent to the curve at x = 1;
   4. the equation of the normal to the curve at x = 1.