Tangents and Normals: The Geometry Hidden in the Derivative
The derivative is the slope of the tangent. With that single fact you can write the equation of a tangent line, the line perpendicular to it (the normal), measure the segments they cut off on the axes, find the angle at which two curves meet, and even handle curves given in polar coordinates. This post collects all these formulas and shows where each one comes from.
1. Equations of the tangent and the normal
Let \(y = f(x)\) be a differentiable function and \(M(x, y)\) a point on its graph. The tangent at \(M\) is the line through \(M\) with slope \(y' = f'(x) = \operatorname{tg}\alpha\) (post on derivatives). The normal at \(M\) is the line through \(M\) perpendicular to the tangent.
Equations
The tangent and the normal to the graph of \(y = f(x)\) at the point \(M(x, y)\) are, respectively,
\[ Y - y = y'\,(X - x) \qquad \text{and} \qquad Y - y = -\frac{1}{y'}\,(X - x), \]
where \(X\) and \(Y\) are the running coordinates of a point on the tangent or normal, and \(y'\) is the value of the derivative of \(y = f(x)\) at \(M\).
Two notational points that confuse many students:
Two pairs of coordinates. Lowercase \((x, y)\) is the fixed point of tangency \(M\); uppercase \((X, Y)\) is an arbitrary point that moves along the line. It is the point-slope form of a line, \( Y - y_0 = k(X - x_0) \), with \(k = y'\).
Why \( -\frac{1}{y'} \)? Two lines with slopes \(k_1\) and \(k_2\) are perpendicular exactly when \( k_1k_2 = -1 \). The normal's slope is therefore the negative reciprocal of the tangent's.
Special cases
If \( y' = 0 \), the tangent is horizontal, \( Y = y \), and the normal is the vertical line \( X = x \) (the formula \(-\frac{1}{y'}\) breaks down, but the geometry is clear).
If \( y' = \pm\infty \) (an infinite derivative, as for \(\sqrt[3]{x}\) at \(0\)), the tangent is vertical, \( X = x \), and the normal is horizontal, \( Y = y \).
Example 1
Tangent and normal to \( y = x^2 \) at \( M(1, 1) \): \( y' = 2x = 2 \).
\[ \text{tangent: } Y - 1 = 2(X - 1) \iff Y = 2X - 1, \qquad \text{normal: } Y - 1 = -\tfrac12(X - 1) \iff Y = -\tfrac12X + \tfrac32. \]
2. Tangent, normal, subtangent, subnormal
The tangent and the normal at \(M\) cut the \(x\)-axis at points \(T\) and \(N\). Drop a perpendicular from \(M\) to the axis, with foot \(P\). This creates four segments with traditional names.
Figure 1. The tangent \(MT\), the normal \(MN\), the subtangent \(PT\) and the subnormal \(PN\) at a point \(M\) of the curve. The angle \(\alpha\) between the tangent and the \(x\)-axis reappears at \(M\), between \(MP\) and the normal \(MN\).
Lengths
The lengths of the tangent, the normal, the subtangent and the subnormal are given by
\[ TM = \left|\frac{y}{y'}\right|\sqrt{1 + y'^2}, \qquad MN = |y|\sqrt{1 + y'^2}, \]
\[ PT = \left|\frac{y}{y'}\right|, \qquad PN = |y\,y'|, \qquad \text{where } y' = \operatorname{tg}\alpha. \]
Where they come from. Everything happens in the right triangle \(TPM\) and its partner \(MPN\), both with the vertical leg \( MP = |y| \).
Subtangent. Setting \(Y = 0\) in the tangent equation gives \( X = x - \frac{y}{y'} \), so \( PT = |x - X| = \left|\frac{y}{y'}\right| \). Equivalently, in triangle \(TPM\): \( \operatorname{tg}\alpha = \frac{MP}{PT} \).
Subnormal. Setting \(Y = 0\) in the normal equation gives \( X = x + y\,y' \), so \( PN = |y\,y'| \). In triangle \(MPN\) the angle at \(M\) is again \(\alpha\) (sides perpendicular to those of the angle at \(T\)), so \( PN = MP\operatorname{tg}\alpha \).
Note that "tangent" and "normal" here mean segments (from \(M\) to the axis), not the whole lines. The formulas with \(y'\) in the denominator need \(y' \ne 0\): a horizontal tangent never reaches the \(x\)-axis (unless it lies on it).
3. Classic examples
Example 2 · The parabola \( y^2 = 4x \) (Figure 1)
For the upper branch \( y = 2\sqrt x \): \( y' = \frac{1}{\sqrt x} \). Then
\[ PT = \left|\frac{y}{y'}\right| = 2\sqrt x\cdot\sqrt x = 2x, \qquad PN = |y\,y'| = 2\sqrt x\cdot\frac{1}{\sqrt x} = 2. \]
Two beautiful facts follow:
The subtangent is \(2x\), so \(T\) lies at \(X = -x\): the vertex bisects the subtangent. To draw the tangent at \(M\) with a ruler, mark the point \(-x\) on the axis and connect it with \(M\).
The subnormal is constant, equal to \(2\) for every point of the parabola. (For \( y^2 = 2px \) it equals \(p\).)
Example 3 · The exponential curve
For \( y = e^x \): \( y' = e^x = y \), so the subtangent is \( PT = \left|\frac{y}{y'}\right| = 1 \) at every point. The tangent to \(e^x\) always hits the \(x\)-axis exactly one unit to the left of the point of tangency. More generally, the curves \( y = Ce^{x/a} \) are exactly the curves with constant subtangent \(a\), which is a first-order differential equation, \( \frac{y}{y'} = a \) (post on differential equations).
Example 4 · The normal to a circle passes through its centre
For \( x^2 + y^2 = r^2 \), implicit differentiation gives \( 2x + 2yy' = 0 \), so \( y' = -\frac xy \). The normal at \(M(x, y)\):
\[ Y - y = \frac yx(X - x) \iff Y = \frac yx\,X. \]
It passes through the origin, the centre of the circle. So a radius is always perpendicular to the tangent, a fact from school geometry, now proved with a derivative.
4. The angle between two curves
The angle between two curves at a common point is defined as the angle between their tangents at that point. If the slopes are \(k_1 = y_1'\) and \(k_2 = y_2'\), the angle \(\theta\) satisfies
If \(k_1k_2 = -1\), the curves are orthogonal (they cross at a right angle); if \(k_1 = k_2\), they touch (have a common tangent). This is the formula behind the isogonal and orthogonal trajectories in the post on differential equations.
Example 5
The curves \( y = x^2 \) and \( y = \sqrt x \) meet at \( (1, 1) \). Slopes: \( k_1 = 2x = 2 \), \( k_2 = \frac{1}{2\sqrt x} = \frac12 \):
\[ \operatorname{tg}\theta = \frac{2 - \frac12}{1 + 2\cdot\frac12} = \frac34, \qquad \theta = \operatorname{arctg}\frac34 \approx 36.9^\circ. \]
The two curves are mirror images in the line \(y = x\), and so are their tangents.
5. Curves in polar coordinates
Many curves (spirals, cardioids, roses) are simplest in polar coordinates, \( \rho = f(\varphi) \), where \(\rho\) is the distance from the pole \(O\) and \(\varphi\) the angle from the polar axis \(OP\). For such curves the natural question is not the slope of the tangent, but the angle it makes with the radius vector \(OM\).
Figure 2. A curve \( \rho = f(\varphi) \) in polar coordinates. The angle \(\beta\) between the radius vector \(OM\) and the tangent \(MT\) is given by \( \operatorname{tg}\beta = \frac{\rho}{\rho'} \). (The curve shown is the logarithmic spiral \( \rho = e^{\varphi} \), for which \(\beta = 45^\circ\) at every point.)
Angle between radius vector and tangent
If \( \rho = f(\varphi) \) is the equation of a curve in polar coordinates and \(\beta\) the angle between the radius vector \(OM\) and the tangent \(MT\), then
\[ \operatorname{tg}\beta = \frac{\rho}{\rho'}, \qquad \rho' = \frac{d\rho}{d\varphi}. \]
Intuition. When the angle grows by a small \(d\varphi\), the point \(M\) moves by \(\rho\,d\varphi\) across the radius (along a small circular arc) and by \(d\rho\) along the radius. These two movements form a tiny right triangle, whose hypotenuse lies along the tangent. The angle opposite the side \(\rho\,d\varphi\) is \(\beta\), so
Exact derivation. With \( x = \rho\cos\varphi \), \( y = \rho\sin\varphi \) (a parametric curve with parameter \(\varphi\)), the tangent makes the angle \(\alpha\) with the \(x\)-axis, where
Since \( \beta = \alpha - \varphi \), the subtraction formula \( \operatorname{tg}(\alpha - \varphi) = \frac{\operatorname{tg}\alpha - \operatorname{tg}\varphi}{1 + \operatorname{tg}\alpha\operatorname{tg}\varphi} \), after simplifying, gives exactly \( \frac{\rho}{\rho'} \).
Example 6 · The logarithmic spiral (Figure 2)
For \( \rho = e^{a\varphi} \): \( \rho' = a\,e^{a\varphi} \), so
\[ \operatorname{tg}\beta = \frac{1}{a} \quad \text{is constant.} \]
The spiral crosses every ray from the pole at the same angle; that's why it is also called the equiangular spiral. It is precisely the isogonal trajectory of the family of rays found in the post on differential equations (there \(a = 1\), \(\beta = 45^\circ\)). Nautilus shells, hurricanes and the approach path of a moth circling a lamp follow this curve.
Example 7 · The Archimedean spiral
For \( \rho = a\varphi \): \( \rho' = a \), so \( \operatorname{tg}\beta = \varphi \). As \(\varphi\) grows, \(\beta \to 90^\circ\): far from the centre, the coils are almost perpendicular to the radius, so they look more and more like circles. (The grooves of a vinyl record follow this spiral.)
Example 8 · The cardioid
For \( \rho = a(1 + \cos\varphi) \): \( \rho' = -a\sin\varphi \), and using the half-angle formulas,
\[ \operatorname{tg}\beta = \frac{1 + \cos\varphi}{-\sin\varphi} = -\frac{2\cos^2\frac\varphi2}{2\sin\frac\varphi2\cos\frac\varphi2} = -\operatorname{ctg}\frac\varphi2 = \operatorname{tg}\Big(\frac\pi2 + \frac\varphi2\Big). \]
So \( \beta = \frac\pi2 + \frac\varphi2 \): the angle between the radius vector and the tangent grows at exactly half the rate of \(\varphi\).
Problem 5: Show that the tangent to the hyperbola \( xy = a^2 \) forms with the coordinate axes a triangle of constant area.
At \( (x_0, \frac{a^2}{x_0}) \): \( y' = -\frac{a^2}{x_0^2} \). The tangent meets the axes at \( X = 2x_0 \) and \( Y = \frac{2a^2}{x_0} \). Area \( = \frac12\cdot 2x_0\cdot\frac{2a^2}{x_0} = 2a^2 \), independent of the point. (Also, \(M\) is the midpoint of the tangent segment between the axes.)
Problem 6: At what angle do the parabolas \( y = x^2 \) and \( y^2 = x \) intersect at the origin?
At \( (0, 0) \) the first has the horizontal tangent \( y = 0 \), the second the vertical tangent \( x = 0 \). They cross at \(90^\circ\).
Problem 7: Find \(\operatorname{tg}\beta\) for the hyperbolic spiral \( \rho = \frac a\varphi \).
Problem 8: For the circle \( \rho = 2a\cos\varphi \), find \(\beta\). What does the result mean geometrically?
\( \rho' = -2a\sin\varphi \), \( \operatorname{tg}\beta = -\operatorname{ctg}\varphi = \operatorname{tg}\big(\frac\pi2 + \varphi\big) \), so \( \beta = \frac\pi2 + \varphi \). This circle passes through the pole; the angle between a chord from the pole and the tangent at its end equals \(\frac\pi2 + \varphi\), in agreement with the inscribed-angle theorem.
Final thoughts
Every formula in this post grows from one fact: the derivative is the slope of the tangent. Add perpendicularity (\(k_1k_2 = -1\)), a right triangle and Pythagoras, and you get the normal and the four classical segments. Switch to polar coordinates, and the same idea, applied to a tiny triangle with sides \(\rho\,d\varphi\) and \(d\rho\), gives the elegant \( \operatorname{tg}\beta = \frac{\rho}{\rho'} \). Geometry and calculus turn out to be two languages for the same picture.