This note shows how to use the derivative to find the equation of the tangent and the normal to a curve at a given point. It also covers questions in which you must first find the point from information about the gradient. These problems join together differentiation and the equation of a straight line.
Key ideas
Tangent
The tangent to a curve at a point is the straight line that touches the curve there and has the same gradient as the curve. If the point is \(x=a\), then the gradient of the tangent is \(m=f'(a)\).
Normal
The normal at a point is the straight line through that point that is perpendicular to the tangent. Two perpendicular lines have gradients whose product is \(-1\), so the gradient of the normal is
\[ m_{\text{normal}}=-\frac{1}{f'(a)}. \]For example, a tangent of gradient 4 has a normal of gradient \(-\tfrac14\). A tangent of gradient \(-\tfrac23\) has a normal of gradient \(\tfrac32\).
Method
- Find the \(y\)-coordinate of the point by substituting \(x=a\) into the equation of the curve, if it is not given.
- Differentiate and substitute \(x=a\) to get the gradient of the tangent.
- Use \(y-y_1=m(x-x_1)\) with the point \((x_1,y_1)\) and the correct gradient. Use \(m=f'(a)\) for the tangent and \(m=-\dfrac{1}{f'(a)}\) for the normal.
- Rearrange into the form the question asks for, such as \(y=mx+c\) or \(ax+by+d=0\) with integer coefficients.
Special cases
- If \(f'(a)=0\), the tangent is horizontal with equation \(y=f(a)\), and the normal is the vertical line \(x=a\). The formula \(-\dfrac{1}{f'(a)}\) cannot be used because it would divide by zero.
- A tangent that is parallel to a given line has the same gradient as that line. So solve \(f'(x)=m\).
- A tangent that is perpendicular to a given line has a gradient that is the negative reciprocal of that line's gradient.
Further steps
Questions often continue after the equation is found. You may need to find where the line meets an axis: put \(y=0\) for the \(x\)-intercept or \(x=0\) for the \(y\)-intercept. You may also need to find where the line meets the curve again, by solving the two equations together. A graphing calculator can help you check a tangent by plotting both curves.
Connecting with the graph
It helps to picture the situation. At a point where the curve is rising steeply, the tangent has a large positive gradient and the normal is nearly flat. At a minimum point, the tangent is horizontal and the normal is vertical. Sketching a quick graph lets you check that the signs of your gradients are sensible, for instance that a falling curve gives a negative tangent gradient and a positive normal gradient.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Tangent gradient \(m=f'(a)\) | The gradient of the curve at \(x=a\). |
| Normal gradient \(-\dfrac{1}{f'(a)}\) | Perpendicular to the tangent, valid when \(f'(a)\ne 0\). |
| \(y-y_1=m(x-x_1)\) | Equation of a line through \((x_1,y_1)\) with gradient \(m\). |
| \(m_1m_2=-1\) | Condition for two lines to be perpendicular. |
| \(m_1=m_2\) | Condition for two lines to be parallel. |
Worked examples
Example 1 The curve \(y=x^3-4x+1\) passes through the point where \(x=2\). Find the equation of the tangent and the equation of the normal at this point. Give the normal in the form \(ax+by+d=0\), where \(a\), \(b\) and \(d\) are integers.
When \(x=2\): \(y=8-8+1=1\). The point is \((2,1)\).
\(\dfrac{dy}{dx}=3x^2-4\). At \(x=2\): \(3(4)-4=8\).
Tangent: \(y-1=8(x-2)\), so \(y=8x-15\).
The normal gradient is \(-\dfrac18\). Normal: \(y-1=-\dfrac18(x-2)\).
Multiply by 8: \(8y-8=-x+2\), so \(x+8y-10=0\).
Answer: tangent \(y=8x-15\); normal \(x+8y-10=0\).
Example 2 Find the equation of the tangent to \(y=x^2-3x+5\) that is parallel to the line \(y=5x+1\).
The gradient of the tangent must be 5.
\(\dfrac{dy}{dx}=2x-3\). Solve \(2x-3=5\): \(x=4\).
When \(x=4\): \(y=16-12+5=9\). The point is \((4,9)\).
Tangent: \(y-9=5(x-4)\), so \(y=5x-11\).
Answer: \(y=5x-11\).
Example 3 The curve \(y=\dfrac{12}{x}\) passes through the point \(P\) where \(x=3\). The normal at \(P\) meets the \(x\)-axis at \(Q\). Find the coordinates of \(Q\).
When \(x=3\): \(y=4\). So \(P=(3,4)\).
Write \(y=12x^{-1}\). Then \(\dfrac{dy}{dx}=-12x^{-2}=-\dfrac{12}{x^2}\).
At \(x=3\): \(-\dfrac{12}{9}=-\dfrac43\). This is the tangent gradient.
The normal gradient is \(-1\div\left(-\dfrac43\right)=\dfrac34\).
Normal: \(y-4=\dfrac34(x-3)\), so \(y=\dfrac34x+\dfrac74\).
Put \(y=0\): \(\dfrac34x=-\dfrac74\), so \(x=-\dfrac73\).
Answer: \(Q=\left(-\dfrac73,\,0\right)\), which is \((-2.33, 0)\) to 3 s.f.
Common errors
- Common error: using the gradient of the tangent for the normal, or the negative reciprocal for the tangent. Read the question carefully for which line it asks for.
- Common error: taking only the negative of the gradient or only the reciprocal. The normal gradient needs both: flip and change the sign.
- Common error: substituting the \(x\)-value into \(y=mx+c\) before finding \(c\), or using the wrong point. Use \(y-y_1=m(x-x_1)\) with the point on the curve.
- Common error: using the \(y\)-value from the line instead of from the curve, or forgetting to calculate \(y\) at all.
- Common error: leaving the answer in the wrong form. If integer coefficients are requested, clear all fractions.
Exam tip: write the point and the gradient on separate lines before you write the equation. Clear working also makes it easy to spot a slip in the gradient.
Check yourself
- Find the equation of the tangent to \(y=x^2\) at the point where \(x=3\).
Answer
\(y=9\) and \(\dfrac{dy}{dx}=2x=6\). Tangent: \(y-9=6(x-3)\), so \(y=6x-9\). - Find the equation of the normal to \(y=x^2\) at the point where \(x=3\), in the form \(ax+by+d=0\).
Answer
The normal gradient is \(-\tfrac16\). \(y-9=-\tfrac16(x-3)\). Multiply by 6: \(6y-54=-x+3\), so \(x+6y-57=0\). - Find where the tangent to \(y=x^2+4\) at \(x=1\) meets the \(x\)-axis.
Answer
\(y=5\), gradient \(=2x=2\). Tangent: \(y-5=2(x-1)\), so \(y=2x+3\). When \(y=0\): \(x=-1.5\). The tangent meets the \(x\)-axis at \((-1.5, 0)\). - The normal to a curve at a point \(P\) has gradient \(-\tfrac14\). What is the gradient of the curve at \(P\)?
Answer
The tangent and normal are perpendicular, so the tangent gradient is the negative reciprocal of \(-\tfrac14\), which is 4.