In short

At any point where a curve y = f(x) is differentiable, the tangent line is the straight line that touches the curve and has slope f'(a), and the normal line is the line perpendicular to the tangent at the same point. The tangent equation is y - f(a) = f'(a)(x - a), and the normal equation is y - f(a) = -\frac{1}{f'(a)}(x - a). Together they let you measure how a curve bends, how two curves meet, and how far the curve stretches along its tangent.

Stand at a bend in a highway and look straight ahead along the road. The direction you are looking is the tangent direction — it is the direction the road is heading at that exact point. If you look sideways, perpendicular to the road, that is the normal direction — the direction the road is curving toward.

Now shrink the highway to a curve on paper and shrink yourself to a point on that curve. The tangent line at that point is the straight line that best approximates the curve right there — it has the same slope as the curve at that instant. The normal line is perpendicular to the tangent, pointing "into" or "out of" the curve's bend.

The derivative gives you the slope of the tangent. From there, the equation of the tangent line is just the point-slope form of a straight line — something you already know from coordinate geometry. The normal follows immediately, because perpendicular lines have slopes that are negative reciprocals.

Equation of the tangent

Take a curve y = f(x) and a point P = (a, f(a)) on it. The slope of the curve at P is f'(a). The tangent line passes through P with that slope.

Equation of the tangent line

At the point (a, f(a)) on the curve y = f(x), the equation of the tangent line is

y - f(a) = f'(a) \cdot (x - a)

This is the point-slope form of a line with slope m = f'(a) passing through (a, f(a)).

Two special cases:

Finding the tangent to a specific curve

For the parabola y = x^2 at the point (3, 9):

The slope is f'(3) = 2(3) = 6. The tangent line is y - 9 = 6(x - 3), which simplifies to y = 6x - 9.

Check: at x = 3, the tangent gives y = 18 - 9 = 9, which matches the curve. And the slope of the tangent (6) matches the derivative (2 \times 3 = 6). The tangent line passes through the right point with the right slope — that is all a tangent needs to do.

The parabola $y = x^2$ and its tangent line $y = 6x - 9$ at the point $(3, 9)$. The tangent touches the parabola at exactly one point and has slope $6 = f'(3)$.

Equation of the normal

The normal line at (a, f(a)) is perpendicular to the tangent. Two perpendicular lines have slopes that are negative reciprocals: if one has slope m, the other has slope -1/m.

Equation of the normal line

At the point (a, f(a)) on the curve y = f(x), with f'(a) \neq 0, the equation of the normal line is

y - f(a) = -\frac{1}{f'(a)} \cdot (x - a)

If f'(a) = 0 (horizontal tangent), the normal is the vertical line x = a.

For the parabola y = x^2 at (3, 9): the tangent has slope 6, so the normal has slope -1/6. The normal equation is y - 9 = -\frac{1}{6}(x - 3), or y = -\frac{x}{6} + \frac{19}{2}.

The tangent and normal are perpendicular at the point of contact and together they form a local coordinate system at that point — one axis along the curve, one across it.

Tangent and normal from an external point

Sometimes you need the tangent to a curve from a point that is not on the curve. For example: find the tangent to y = x^2 from the point (0, -1).

The idea: suppose the tangent touches the curve at some unknown point (t, t^2). The tangent line at that point is y - t^2 = 2t(x - t), which simplifies to y = 2tx - t^2. This line must pass through (0, -1):

-1 = 2t(0) - t^2 = -t^2 \implies t^2 = 1 \implies t = \pm 1

So there are two tangent lines from (0, -1) to the parabola: y = 2x - 1 (touching at (1, 1)) and y = -2x - 1 (touching at (-1, 1)).

The general method: parametrise the contact point as (t, f(t)), write the tangent line equation, impose the condition that it passes through the external point, and solve for t.

Two tangent lines to $y = x^2$ drawn from the external point $(0, -1)$. The tangents touch the parabola at $(1, 1)$ and $(-1, 1)$. From a point below a parabola, there are always exactly two tangent lines (as long as the point is outside the parabola).

Angle of intersection of two curves

When two curves cross, they form an angle at the crossing point. The angle of intersection is defined as the acute angle between their tangent lines at that point.

If curve y = f(x) has slope m_1 = f'(a) and curve y = g(x) has slope m_2 = g'(a) at the point of intersection (a, b), then the tangent of the angle \theta between the tangent lines is

\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|

This is the formula for the angle between two lines, applied to the two tangent lines. The absolute value ensures \theta is acute (between 0 and \pi/2).

Special cases:

Finding where two curves meet

To use the angle formula, you first need the intersection point. Solve f(x) = g(x) to find x, then compute f'(x) and g'(x) at that point.

Take y = x^2 and y = x^3. Set x^2 = x^3: x^2(1 - x) = 0, so x = 0 or x = 1.

At x = 1: m_1 = 2(1) = 2 and m_2 = 3(1)^2 = 3. The angle is \tan\theta = \left|\frac{2 - 3}{1 + 6}\right| = \frac{1}{7}, so \theta = \tan^{-1}(1/7) \approx 8.13°.

At x = 0: m_1 = 0 and m_2 = 0. Both tangent lines are horizontal — the curves are tangent to each other at the origin. The angle is .

The curves $y = x^2$ (black) and $y = x^3$ (red) intersect at $(0, 0)$ and $(1, 1)$. At the origin, both curves have slope $0$, so they are tangent to each other. At $(1, 1)$, the tangent lines have slopes $2$ and $3$, making an angle of about $8.1°$.

Lengths: tangent, normal, subtangent, subnormal

At a point P = (a, f(a)) on a curve, the tangent and normal lines each hit the x-axis at some point. The distances between P, these intersection points, and the foot of the perpendicular from P to the x-axis define four geometric lengths.

Let m = f'(a) and k = f(a) (the y-coordinate of P). Drop a perpendicular from P to the x-axis; its foot is M = (a, 0).

Tangent, normal, subtangent, and subnormal at a point on a curveA curve with a point P on it. The tangent line from P meets the x-axis at T. The normal line from P meets the x-axis at N. The perpendicular from P meets the x-axis at M. The segment TM is the subtangent, MN is the subnormal, PT is the length of tangent, and PN is the length of normal. P(a, f(a)) M T N subtangent subnormal
The geometry of tangent and normal at a point $P$ on a curve. The tangent meets the $x$-axis at $T$; the normal meets it at $N$; the perpendicular foot is $M$. The four lengths are: $PT$ (length of tangent), $PN$ (length of normal), $TM$ (subtangent), and $MN$ (subnormal).

Finding the four lengths

The tangent line at P = (a, k) is y - k = m(x - a). It meets the x-axis (y = 0) at the point T:

0 - k = m(x - a) \implies x = a - \frac{k}{m}

So T = \left(a - \frac{k}{m}, \, 0\right).

The normal line at P is y - k = -\frac{1}{m}(x - a). It meets the x-axis at N:

0 - k = -\frac{1}{m}(x - a) \implies x = a + km

So N = (a + km, \, 0).

The four lengths:

Lengths of tangent, normal, subtangent, subnormal

At a point (a, f(a)) on y = f(x), with m = f'(a) and k = f(a):

Subtangent = TM = \left|\dfrac{k}{m}\right|

Subnormal = MN = |k \cdot m|

Length of tangent = PT = |k| \cdot \sqrt{1 + \dfrac{1}{m^2}} = \dfrac{|k|}{|m|}\sqrt{1 + m^2}

Length of normal = PN = |k| \cdot \sqrt{1 + m^2}

The subtangent and subnormal are measured along the x-axis. The length of tangent and length of normal are measured along the tangent and normal lines themselves (from P to where they meet the axis).

Notice that the subnormal of the parabola y = ax^2 at any point is |y \cdot 2ax| = |ax^2 \cdot 2ax| = 2a^2|x|^3... but for the standard parabola y^2 = 4ax, the subnormal is constant. This is a famous property: for a parabola y^2 = 4ax, implicit differentiation gives 2y \cdot y' = 4a, so y' = 2a/y. The subnormal is |y \cdot y'| = |y \cdot 2a/y| = 2a — independent of the point. The subnormal of a parabola is always equal to the semi-latus rectum.

Computing the tangent and normal together

Example 1: Tangent and normal to $y = \sqrt{x}$ at $(4, 2)$

Step 1. Find the derivative.

f(x) = x^{1/2} \implies f'(x) = \frac{1}{2\sqrt{x}}

At x = 4: f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}.

Why: the power rule gives \frac{d}{dx}x^{1/2} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}.

Step 2. Write the tangent equation.

y - 2 = \frac{1}{4}(x - 4) \implies y = \frac{x}{4} + 1

Why: point-slope form with slope m = 1/4 and point (4, 2).

Step 3. Write the normal equation.

The normal slope is -1/m = -4.

y - 2 = -4(x - 4) \implies y = -4x + 18

Why: perpendicular lines have slopes that are negative reciprocals. If the tangent slope is 1/4, the normal slope is -4.

Step 4. Find the four lengths.

With k = 2 and m = 1/4:

  • Subtangent = |k/m| = |2/(1/4)| = 8
  • Subnormal = |km| = |2 \cdot 1/4| = 1/2
  • Length of tangent = \frac{|k|}{|m|}\sqrt{1 + m^2} = \frac{2}{1/4}\sqrt{1 + 1/16} = 8 \cdot \frac{\sqrt{17}}{4} = 2\sqrt{17}
  • Length of normal = |k|\sqrt{1 + m^2} = 2\sqrt{1 + 1/16} = 2 \cdot \frac{\sqrt{17}}{4} = \frac{\sqrt{17}}{2}

Why: substituting directly into the formulas. The subtangent is much larger than the subnormal because the slope is small — the tangent line is nearly flat, so it travels a long way along the x-axis before reaching P.

Result: Tangent: y = \frac{x}{4} + 1. Normal: y = -4x + 18. Subtangent = 8. Subnormal = \frac{1}{2}.

The curve $y = \sqrt{x}$ with its tangent (red, slope $1/4$) and normal (soft red, slope $-4$) at the point $P = (4, 2)$. The tangent is nearly flat; the normal is steep. The tangent meets the $x$-axis at $T = (-4, 0)$, making the subtangent $TM = 8$ units long. The normal meets the axis at $N = (4.5, 0)$, making the subnormal just $MN = 0.5$ units.

The picture shows the asymmetry vividly. Because the slope at (4, 2) is gentle (only 1/4), the tangent extends far to the left before hitting the axis, while the steep normal plunges down almost vertically and meets the axis right next to M.

Example 2: Angle of intersection of $y = x^2$ and $y = x$

Step 1. Find the intersection points.

x^2 = x \implies x^2 - x = 0 \implies x(x - 1) = 0 \implies x = 0 \text{ or } x = 1

The intersection points are (0, 0) and (1, 1).

Why: set the two expressions for y equal and solve for x.

Step 2. Compute the slopes at (1, 1).

For y = x^2: m_1 = 2x = 2(1) = 2.

For y = x: m_2 = 1.

Why: differentiate each curve and evaluate at the intersection point.

Step 3. Apply the angle formula.

\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| = \left|\frac{2 - 1}{1 + 2 \cdot 1}\right| = \frac{1}{3}
\theta = \tan^{-1}\!\left(\frac{1}{3}\right) \approx 18.43°

Why: the formula gives the tangent of the acute angle between the two tangent lines at the intersection point.

Step 4. Check the other intersection point (0, 0).

m_1 = 2(0) = 0 and m_2 = 1. So \tan\theta = \left|\frac{0 - 1}{1 + 0}\right| = 1, giving \theta = 45°.

Why: at the origin, y = x^2 is flat (horizontal tangent) while y = x has slope 1. The angle between a horizontal line and a 45° line is 45°.

Result: The curves intersect at angles 45° (at the origin) and \tan^{-1}(1/3) \approx 18.43° (at (1, 1)).

The parabola $y = x^2$ and the line $y = x$ intersect at $(0, 0)$ and $(1, 1)$. At the origin, the parabola's tangent is horizontal, so the two curves meet at $45°$. At $(1, 1)$, the parabola's tangent has slope $2$ while the line has slope $1$, making a smaller angle of about $18.4°$.

The graph confirms both results. At the origin, the line y = x clearly cuts across the flat bottom of the parabola at a wide angle. At (1, 1), the parabola has steepened toward the line's slope, so they cross at a much narrower angle.

Common confusions

Going deeper

If you came here to find tangent and normal equations and compute intersection angles, you have everything you need — stop here. The rest is for readers who want the geometry behind the formulas.

Why the subtangent matters

The subtangent |f(a)/f'(a)| has a clean geometric meaning: it tells you how far back you would need to start on the x-axis to "reach" the point P by walking up the tangent line. For exponential curves, the subtangent is constant. Consider y = e^x: f'(a) = e^a = f(a), so the subtangent is |f(a)/f'(a)| = 1 at every point. The tangent to e^x always crosses the x-axis exactly 1 unit to the left of the point of tangency. This is a defining property of the exponential function.

For the logarithmic curve y = \ln x: f'(a) = 1/a, so the subtangent is |f(a)/f'(a)| = |a \ln a|, which grows without bound. As a \to \infty, the subtangent grows faster than a — the tangent has to reach further and further back to hit the axis.

Orthogonal trajectories

Two families of curves are orthogonal trajectories of each other if every curve in one family intersects every curve in the other family at right angles. For example, the family of concentric circles x^2 + y^2 = r^2 and the family of straight lines y = mx through the origin are orthogonal trajectories — every radius of a circle is perpendicular to the circle at the point of intersection.

To find orthogonal trajectories of a family F(x, y, c) = 0:

  1. Differentiate to get a relation between x, y, and dy/dx (eliminate c).
  2. Replace dy/dx with -dx/dy (swap the slope with its negative reciprocal).
  3. Solve the resulting differential equation.

The condition m_1 \cdot m_2 = -1 — the angle-of-intersection formula with \theta = 90° — is what drives the entire theory of orthogonal trajectories. It appears in physics (electric field lines and equipotential surfaces), fluid dynamics (streamlines and potential lines), and conformal mapping in complex analysis.

Tangent and normal in parametric form

If a curve is given parametrically as x = \phi(t), y = \psi(t), the slope at a point is \frac{dy}{dx} = \frac{\psi'(t)}{\phi'(t)}. The tangent at the point (\phi(t_0), \psi(t_0)) is

\frac{y - \psi(t_0)}{\psi'(t_0)} = \frac{x - \phi(t_0)}{\phi'(t_0)}

and the normal is

\frac{y - \psi(t_0)}{-\phi'(t_0)} = \frac{x - \phi(t_0)}{\psi'(t_0)}

For a circle x = r\cos t, y = r\sin t: \frac{dy}{dx} = \frac{r\cos t}{-r\sin t} = -\cot t. The normal slope is \tan t. The normal line passes through (r\cos t, r\sin t) with slope \tan t — and if you extend it, it passes through the origin. The normal to a circle always points toward (or away from) the centre. That is a geometric fact you know already, but here it emerges from the calculus.

Where this leads next