Tangents and normals

A curve is not a straight line, but at any single point it has a best straight-line imitation: the tangent, the line that just grazes the curve there without crossing through it locally. Perpendicular to that tangent, through the same point, sits the normal. Both lines come from one number — the derivative $\dfrac{dy}{dx}$ evaluated at the point — combined with two pieces of coordinate geometry you already know: the point-slope form of a line, and the fact that perpendicular slopes are negative reciprocals of each other. This topic, the geometric face of differentiation, sits inside "Applications of Derivatives," the sub-unit that follows on directly from the chain rule, implicit differentiation, parametric differentiation and logarithmic differentiation you have already built up. It rewards anyone who can differentiate correctly and then apply two short formulas — point-slope form and the perpendicular-slopes condition — without mixing them up, which is exactly why these questions are a favourite way for examiners to test the whole differentiation toolkit in one short problem.

The core insight

The tangent at $(x_1,y_1)$ has slope $m=\left.\dfrac{dy}{dx}\right|_{(x_1,y_1)}$ and equation $y-y_1=m(x-x_1)$ , while the normal is perpendicular to it with slope $-1/m$ .

The core idea

Step 1 — get a numerical slope. Differentiate $y=f(x)$ to get $\dfrac{dy}{dx}$ as a function of $x$ , then substitute the point's $x$ -coordinate to turn it into a single number, the tangent's slope at that exact point. This works no matter how the curve is presented — explicitly as $y=f(x)$ , implicitly as $F(x,y)=0$ (differentiate term by term and solve for $\dfrac{dy}{dx}$ ), or parametrically as $x=x(t),\,y=y(t)$ (use $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ ). Once you have the number, everything downstream is identical.

Step 2 — write the tangent. The tangent line passes through $(x_1,y_1)$ with that slope, so ordinary point-slope form gives it directly:

y - y_1 = m_{\text{tangent}}(x - x_1)

where: $(x_1,y_1)$ is the point of tangency (it must actually lie on the curve) and $m_{\text{tangent}}=\left.\dfrac{dy}{dx}\right|_{(x_1,y_1)}$ .

Step 3 — flip to the normal. Because the normal is perpendicular to the tangent at the same point, its slope is the negative reciprocal:

m_{\text{normal}} = \frac{-1}{m_{\text{tangent}}}, \qquad y - y_1 = m_{\text{normal}}(x - x_1) \quad (m_{\text{tangent}} \ne 0)

where: the perpendicularity condition $m_1 m_2 = -1$ for two non-vertical perpendicular lines is the same one used throughout coordinate geometry.

Special cases. If $m_{\text{tangent}}=0$ (a horizontal tangent), the tangent is $y=y_1$ and the normal is the vertical line $x=x_1$ — do not attempt " $-1/0$ ." If $\dfrac{dy}{dx}$ is undefined at the point (a vertical tangent), the roles swap: the tangent is $x=x_1$ and the normal is $y=y_1$ .

Figure

A curve with a straight tangent line touching it at one point and a perpendicular normal line crossing through the same point

The extension to implicit and parametric curves changes only how you get the numerical slope in Step 1 — Steps 2 and 3 never change. This is why tangent/normal problems are treated as a cumulative check on the whole differentiation toolkit: get the derivative right, and the geometry is routine. It also helps to remember why the derivative is the right slope to use at all. As a second point on the curve slides closer and closer to $(x_1,y_1)$ , the secant line joining them rotates until, in the limit, it settles into the unique line that best hugs the curve at that instant — and that limiting slope is precisely $\dfrac{dy}{dx}$ evaluated there. The tangent is therefore the best straight-line approximation to the curve near the point, which is also why tangent-slope questions and rate-of-change questions in other parts of the syllabus feel closely related — both read the same first derivative, just for different purposes. The normal carries no such approximation meaning; its role is purely geometric, and it shows up in problems about the shortest distance from a point to a curve, reflection off a curved surface, or families of curves that cross a given curve at right angles.

Derivation ATangent and normal equations from the perpendicularity condition

The goal is to derive, from only the point-slope form of a line and the perpendicular-slopes condition $m_1m_2=-1$ , the two standard equations used for every tangent/normal problem on this syllabus, so that the formulas are not memorised blindly but seen to follow from coordinate geometry already known.

Step 1. Start from point-slope form. Any non-vertical line through a fixed point $(x_1,y_1)$ with slope $m$ satisfies $y-y_1=m(x-x_1)$ — this is a general fact about straight lines, true for any $m$ and any point, independent of calculus.

Step 2. Supply the slope from calculus. For the tangent to $y=f(x)$ at $(x_1,y_1)$ , the only slope that makes the line touch the curve without crossing it locally is the instantaneous rate of change there, $m_{\text{tangent}}=f'(x_1)=\left.\dfrac{dy}{dx}\right|_{(x_1,y_1)}$ . Substituting this specific $m$ into the point-slope form of Step 1 gives the tangent equation $y-y_1=f'(x_1)(x-x_1)$ .

Step 3. Bring in perpendicularity for the normal. Two non-vertical lines are perpendicular exactly when the product of their slopes is $-1$ . Since the normal is defined as the line through $(x_1,y_1)$ perpendicular to the tangent, its slope $m_{\text{normal}}$ must satisfy $m_{\text{tangent}}\cdot m_{\text{normal}}=-1$ , which rearranges to $m_{\text{normal}}=-1/m_{\text{tangent}}$ .

Step 4. Assemble the normal equation. Substituting this slope back into the same point-slope form from Step 1 (still through the same point $(x_1,y_1)$ , since both lines pass through the point of tangency) gives $y-y_1=\dfrac{-1}{f'(x_1)}(x-x_1)$ , valid whenever $f'(x_1)\ne0$ . The two boxed results — $y-y_1=m_{\text{tangent}}(x-x_1)$ and $y-y_1=\dfrac{-1}{m_{\text{tangent}}}(x-x_1)$ — are therefore nothing but point-slope form fed two different, related slopes. Both equations reduce to the special-case lines $y=y_1$ and $x=x_1$ exactly when $m_{\text{tangent}}=0$ or is undefined, consistent with the core-idea discussion above. Notice, too, that nothing in this derivation cared how $f'(x_1)$ was obtained — direct differentiation, implicit differentiation, or the parametric ratio $\dfrac{dy/dt}{dx/dt}$ all hand you the same single number $m_{\text{tangent}}$ , and from that point on Steps 1, 3 and 4 run identically regardless of the curve's original form.

Worked example

Find the equations of the tangent and normal to $y=x^2-3x+2$ at the point $(2,0)$ .

Step 1. Confirm the point lies on the curve: at $x=2$ , $y=4-6+2=0$ , matching $(2,0)$ .

Step 2. Differentiate: $\dfrac{dy}{dx}=2x-3$ .

Step 3. Evaluate the slope at $x=2$ : $m_{\text{tangent}}=2(2)-3=1$ .

Step 4. Write the tangent using point-slope form: $y-0=1\cdot(x-2)$ , which simplifies to $x-y-2=0$ .

Step 5. Find the normal's slope: $m_{\text{normal}}=-1/1=-1$ .

Step 6. Write the normal: $y-0=-1(x-2)$ , which simplifies to $x+y-2=0$ .

Final answer. Tangent: $\boxed{x-y-2=0}$ ; Normal: $\boxed{x+y-2=0}$ . As a quick sanity check, multiply the two slopes: $m_{\text{tangent}}\cdot m_{\text{normal}}=1\times(-1)=-1$ , confirming the perpendicularity condition, and both lines pass through $(2,0)$ by construction — a fast way to catch an arithmetic slip before it costs marks.

Competitive stretch

The same curve $x=at^2,\,y=2at$ from the parametric case gives, at $t=1$ , the point $(a,2a)$ with $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{2a}{2at}=\dfrac1t=1$ , so $m_{\text{tangent}}=1$ and $m_{\text{normal}}=-1$ . Point-slope form at $(a,2a)$ gives $y-2a=-1(x-a)$ , which simplifies to $\boxed{x+y=3a}$ — the same mechanics as above, only the slope now came from a parametric ratio instead of a direct derivative.

Common traps

Using the same slope for both lines — forgetting that the normal needs the negative reciprocal, not $m_{\text{tangent}}$ itself.
Leaving $\dfrac{dy}{dx}$ as a general expression in $x$ and plugging it straight into the line equation instead of first substituting $x_1$ to get a single number; in parametric problems this shows up as substituting the parameter $t$ before $\dfrac{dy}{dx}$ has actually been converted into a pure ratio.
Writing " $m_{\text{normal}}=-1/0$ " when $m_{\text{tangent}}=0$ , instead of recognising the normal is simply the vertical line $x=x_1$ .
Transposing coordinates in point-slope form, e.g. $y-x_1=m(x-y_1)$ instead of $y-y_1=m(x-x_1)$ — a labelling slip that produces an entirely wrong line.

Ask the tutor: want a walkthrough of a tangent/normal problem on an implicit curve like $x^2+y^2=25$ , or a check of your own worked answer?

Do I always need to check the point lies on the curve before finding the tangent?

Yes — the point-slope construction assumes $(x_1,y_1)$ is genuinely on the curve, since the tangent must touch the curve there. If the given point does not satisfy $y=f(x)$ , the derivative you compute is still a valid slope somewhere on the curve, but the line you write down will not actually be tangent to the curve at that point. A quick substitution check, as in Step 1 of the worked example, avoids building an equation for a point that was never on the curve.

Why does the normal use the negative reciprocal instead of just minus the tangent slope?

Perpendicularity between two non-vertical lines requires the product of their slopes to equal $-1$ , not that one slope is the negative of the other. If the tangent has slope $m$ , the normal's slope must satisfy $m \cdot m_{\text{normal}} = -1$ , giving $m_{\text{normal}} = -1/m$ . Confusing "negative reciprocal" with "negative" is one of the most common slips — always divide $-1$ by $m$ , don't just flip the sign.

How is the parametric case different from the explicit case once I have the slope?

It isn't, once you reach a numerical slope. The only extra work in the parametric case is computing $\dfrac{dy}{dx}$ as $\dfrac{dy/dt}{dx/dt}$ instead of differentiating $y$ directly with respect to $x$ , and substituting the specific parameter value (not $x_1$ directly) into that ratio. After that number is in hand, the point-slope construction for the tangent and the negative-reciprocal construction for the normal are exactly the same two steps used for an explicit curve.

Tangents and normals

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads