Tangents and Normals

The tangent as the limiting secant. For a curve $y = f(x)$, the tangent at a point $P(x_0, y_0)$ on the curve is the straight line through $P$ whose slope equals $f'(x_0)$ -- the limiting position of a secant line through $P$ and a second nearby point on the curve, as that second point slides in toward $P$ (this is precisely how the derivative was itself originally defined). Once the slope is known, the tangent's equation follows from the point-slope form of a line.

The normal's slope is $-1/f'(x_0)$ because two lines are perpendicular exactly when the product of their slopes is $-1$ (coordinate-geometry fact, Class XI). Both lines pass through the same point $(x_0, y_0)$ on the curve; only their slopes differ.

Two edge cases, worth stating explicitly.

• If $f'(x_0) = 0$, the tangent is horizontal ($y = y_0$) and the normal is vertical ($x = x_0$), since a slope of $-1/0$ is undefined for a line perpendicular to a horizontal one (see Example 6 and Exercise: Tangents and Normals, Q5).
• If $f'(x_0)$ itself does not exist but the curve has a vertical tangent at $(x_0, y_0)$, the tangent is $x = x_0$ and the normal is $y = y_0$ -- the mirror image of the case above.

Method for problems phrased in terms of a given direction. Many problems do not name the point directly but instead specify a direction the tangent or normal must have -- e.g. "parallel to the line $4x - 2y + 5 = 0$," or "parallel to the $x$-axis." The method is always:

1. Find the target slope $m$ from the given line (rearrange to $y = mx + c$ form), or $m = 0$ for "parallel to the $x$-axis."
2. If the tangent must have slope $m$, solve $f'(x) = m$ for $x$ (there may be more than one solution -- Exercise: Tangents and Normals, Q3 has two).
3. If instead the normal must have slope $m$ (as in the same Q3), first find the required tangent slope $-1/m$ (perpendicularity), then solve $f'(x) = -1/m$.
4. Find $y$ from the curve's equation at each such $x$, then write the tangent/normal equation at each resulting point using the formulas above. …