In short

A tangent from an external point, the angle at which two curves cross, the conditions for orthogonality, and common tangents to a pair of curves — these are the natural next questions after "find the tangent at a given point." Each one reduces to setting up the right equation and solving it with algebra you already know.

Two parabolas, y = x^2 and y = (x - 3)^2, sit side by side on the plane. The first opens upward from the origin; the second opens upward from (3, 0). Somewhere between them, you can draw a single straight line that is tangent to both curves at once — touching each one without cutting through it.

Where is that line? What is its equation? And how do you find it systematically, instead of by guessing?

These are the questions this article answers. You already know how to write down the tangent to a single curve at a single point. The advanced problems are: tangents from a point that is not on the curve, tangents that are common to two curves, and the angle at which two curves meet when they cross. Each reduces to the same core move — writing down the tangent equation with an unknown parameter and then solving for the parameter — but the setups differ in instructive ways.

Tangent from an external point

When a point P lies on a curve, the tangent at P is unique (assuming the curve is smooth there). But when P lies off the curve, multiple tangent lines from P can touch the curve at different points.

Take the parabola y = x^2 and the external point P = (0, -1). The point P is below the vertex — it is not on the parabola. How many tangent lines to the parabola pass through P?

The method. Let the tangent touch the parabola at the point (a, a^2) for some unknown a. The slope of the parabola at x = a is \frac{dy}{dx} = 2a. So the tangent line at (a, a^2) is

y - a^2 = 2a(x - a)

which simplifies to

y = 2ax - a^2

This tangent line must pass through P = (0, -1). Substitute x = 0, y = -1:

-1 = 2a(0) - a^2 = -a^2

So a^2 = 1, giving a = 1 or a = -1. There are two tangent lines from (0, -1) to the parabola.

When a = 1: the tangent is y = 2x - 1, touching the parabola at (1, 1).

When a = -1: the tangent is y = -2x - 1, touching the parabola at (-1, 1).

Two tangent lines from the external point $P = (0, -1)$ to the parabola $y = x^2$. The tangents touch the curve at $(1,1)$ and $(-1,1)$ — the symmetry around the $y$-axis reflects the symmetry of the parabola.

The key insight is the unknown parameter a. You do not know where the tangent touches the curve, so you let that point be a variable. The tangent equation, expressed in terms of a, gives you a family of lines. Requiring the external point to lie on one of those lines produces an equation in a alone.

How many tangent lines? The number of tangent lines from an external point equals the number of real solutions of that equation. For a parabola, it is a quadratic in a, so at most two tangent lines. For higher-degree curves, you can get more.

Length of tangent, normal, subtangent, and subnormal

At a point P on a curve, the tangent and normal lines meet the x-axis at specific points. The distances involved have classical names that show up in competitive exams.

Let P = (x_0, y_0) lie on the curve y = f(x), with m = f'(x_0) being the slope of the tangent.

The tangent line at P has equation Y - y_0 = m(X - x_0). Setting Y = 0 gives the foot of the tangent on the x-axis:

X_T = x_0 - \frac{y_0}{m}

The normal line at P has equation Y - y_0 = -\frac{1}{m}(X - x_0). Setting Y = 0 gives the foot of the normal on the x-axis:

X_N = x_0 + y_0 m

Drop a perpendicular from P to the x-axis; its foot is M = (x_0, 0). Then:

Lengths associated with tangent and normal

  • Subtangent = |X_T - x_0| = \left|\dfrac{y_0}{m}\right| — the horizontal distance from M to the foot of the tangent.
  • Subnormal = |X_N - x_0| = |y_0 \cdot m| — the horizontal distance from M to the foot of the normal.
  • Length of tangent = |PT| = |y_0| \sqrt{1 + \dfrac{1}{m^2}} — the actual distance from P to the foot of the tangent.
  • Length of normal = |PN| = |y_0| \sqrt{1 + m^2} — the actual distance from P to the foot of the normal.

These follow from elementary coordinate geometry. The subtangent and subnormal are just projections onto the x-axis; the "lengths" are the hypotenuses of the right triangles formed by P, M, and the respective foot.

A useful fact about the parabola. For y^2 = 4ax (the standard parabola), the subnormal is constant: it equals 2a, regardless of where you are on the curve. You can verify this directly. At the point (at^2, 2at), the slope is dy/dx = 1/t, and the subnormal is |y \cdot m| = |2at \cdot (1/t)| = 2a.

Tangent, normal, subtangent, and subnormal at a point on a curveA curve with a point P on it. The tangent and normal at P extend to meet the x-axis at points T and N respectively. The foot of the perpendicular from P to the x-axis is M. The subtangent is the distance TM, the subnormal is MN, the length of tangent is PT, and the length of normal is PN. P(x₀, y₀) T M N subtangent subnormal
The tangent at $P$ meets the $x$-axis at $T$; the normal meets it at $N$. The perpendicular from $P$ to the $x$-axis has foot $M$. The subtangent is $TM$, the subnormal is $MN$, the length of tangent is $PT$, and the length of normal is $PN$.

Angle of intersection of two curves

When two curves cross at a point, the angle of intersection is defined as the acute angle between their tangent lines at that point.

Suppose y = f(x) and y = g(x) meet at a point (x_0, y_0). The slopes of their tangents there are m_1 = f'(x_0) and m_2 = g'(x_0). The angle \theta between the two tangent lines satisfies

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

provided neither tangent is vertical and 1 + m_1 m_2 \neq 0. This is the standard formula for the angle between two lines with known slopes.

Special case: \theta = 90°. When the two tangent lines are perpendicular, \tan\theta is undefined and 1 + m_1 m_2 = 0, i.e., m_1 m_2 = -1. Curves that meet at right angles are called orthogonal at that point.

Special case: \theta = 0°. When the tangent lines are parallel (or identical), m_1 = m_2 and \tan\theta = 0. If the curves actually share the same tangent line, they are said to touch each other (or be tangent to each other).

Angle of intersection

If two curves y = f(x) and y = g(x) intersect at a point where their slopes are m_1 and m_2, then the acute angle \theta between the curves at that point is

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

The curves are orthogonal at the point if m_1 m_2 = -1.

Worked examples

Example 1: Find the angle of intersection of y = x² and y = x³

Step 1. Find the intersection points. Set x^2 = x^3, so x^3 - x^2 = 0, giving x^2(x - 1) = 0. The curves meet at x = 0 and x = 1.

Why: the angle of intersection is defined at each intersection point separately, so you need to find all of them first.

Step 2. Compute slopes at each intersection point. For y = x^2: m_1 = 2x. For y = x^3: m_2 = 3x^2.

At x = 0: m_1 = 0 and m_2 = 0. Both tangent lines are horizontal — they have the same slope. The angle of intersection at the origin is \theta = 0°.

Why: when both slopes are equal, \tan\theta = |0/(1+0)| = 0.

Step 3. Compute the angle at x = 1. Here m_1 = 2 and m_2 = 3.

\tan\theta = \left|\frac{2 - 3}{1 + 2 \cdot 3}\right| = \left|\frac{-1}{7}\right| = \frac{1}{7}

Why: direct substitution into the angle formula. The absolute value makes the angle acute.

Step 4. Convert to an angle.

\theta = \tan^{-1}\left(\frac{1}{7}\right) \approx 8.13°

Why: for JEE-style problems, leaving the answer as \tan^{-1}(1/7) is standard. The decimal value shows the angle is quite small — the curves are nearly parallel at (1,1).

Result: At the origin, the curves touch (\theta = 0°). At (1, 1), they intersect at \theta = \tan^{-1}(1/7) \approx 8.13°.

The parabola $y = x^2$ (black) and the cubic $y = x^3$ (red) intersect at $O$ and $(1,1)$. At the origin both tangents are horizontal — the curves touch. At $(1,1)$ the tangent lines have slopes 2 and 3 — nearly parallel, intersecting at about 8°.

The small angle at (1, 1) matches the visual: the two curves run almost alongside each other near that point.

Example 2: Find the common tangent to y = x² and y = (x − 3)²

Step 1. Write the tangent to the first parabola at a general point. At (a, a^2) on y = x^2, the slope is 2a and the tangent line is

y = 2ax - a^2

Why: this is the parametric form of the tangent, with a as the unknown parameter.

Step 2. Write the tangent to the second parabola at a general point. At (b, (b-3)^2) on y = (x-3)^2, the slope is 2(b-3) and the tangent line is

y = 2(b-3)x - (b-3)^2 - 6(b-3) = 2(b-3)x - b^2 + 3b - 3 \cdot 2 + 9

Expanding more carefully: the tangent at (b, (b-3)^2) is y - (b-3)^2 = 2(b-3)(x - b), which gives

y = 2(b-3)x - 2b(b-3) + (b-3)^2 = 2(b-3)x - (b-3)(2b - b + 3) = 2(b-3)x - (b-3)(b+3)

So y = 2(b-3)x - (b^2 - 9).

Why: applying the same parametric tangent method to the second curve, keeping everything in terms of b.

Step 3. For a common tangent, the two tangent lines must be the same line. That means equal slopes and equal intercepts:

Slopes: 2a = 2(b-3), so a = b - 3.

Intercepts: -a^2 = -(b^2 - 9), so a^2 = b^2 - 9.

Why: two linear equations y = mx + c describe the same line only when their slopes and y-intercepts both match.

Step 4. Substitute a = b - 3 into a^2 = b^2 - 9:

(b-3)^2 = b^2 - 9
b^2 - 6b + 9 = b^2 - 9
-6b = -18
b = 3, \quad a = 0

Why: the quadratic terms cancel, leaving a clean linear equation — one common tangent.

Step 5. Write the common tangent. Using a = 0: y = 2(0)x - 0^2 = 0. The common tangent is y = 0 — the x-axis itself.

Result: The common tangent to y = x^2 and y = (x-3)^2 is the line y = 0. It touches the first parabola at the origin and the second at (3, 0).

The two parabolas $y = x^2$ and $y = (x-3)^2$ share a single common tangent — the $x$-axis. It touches each parabola at its vertex.

The x-axis is tangent to both parabolas at their respective vertices. This makes geometric sense: both parabolas open upward and their vertices sit on the x-axis, so the axis just touches each one from below.

Common tangents — the general strategy

The method in Example 2 generalises. To find common tangents to two curves C_1 and C_2:

  1. Write the tangent to C_1 at a general point, parametrised by some variable a.
  2. Write the tangent to C_2 at a general point, parametrised by b.
  3. Set the slope and intercept of the first equal to the slope and intercept of the second.
  4. Solve the resulting system in a and b.

Each solution gives one common tangent. The number of common tangents depends on the curves — two parabolas opening the same way might have one or two; a line and a circle can have zero, one, or two.

An alternate approach. Sometimes it is faster to write the tangent to one curve and then require it to be tangent to the second. If the tangent to C_1 at parameter a is y = mx + c (with m and c expressed in terms of a), substitute this line into the equation of C_2. The line is tangent to C_2 precisely when the resulting equation has a repeated root — i.e., its discriminant is zero. This gives an equation in a alone.

Condition for a line to be tangent

A line y = mx + c is tangent to a curve y = f(x) if and only if the equation f(x) = mx + c has a repeated root. Equivalently, the line touches the curve at exactly one point (counting multiplicity).

For the parabola y = x^2, the condition is: x^2 = mx + c must have a repeated root, so its discriminant m^2 + 4c = 0, giving c = -m^2/4. Every tangent to y = x^2 has the form y = mx - m^2/4.

For the circle x^2 + y^2 = r^2, substituting y = mx + c gives x^2(1+m^2) + 2mcx + c^2 - r^2 = 0. The discriminant-zero condition yields c^2 = r^2(1 + m^2), i.e., c = \pm r\sqrt{1+m^2}. This is the condition for a line to be tangent to a circle — a result you will use often.

Orthogonal curves

Two families of curves are orthogonal families if every curve of one family intersects every curve of the other at right angles. The classic example is the family of concentric circles x^2 + y^2 = r^2 and the family of straight lines through the origin y = kx. At every point of intersection, the radius and the line through the origin coincide, while the circle's tangent is perpendicular to the radius — so the curves meet at 90°.

To show that two families are orthogonal, the standard method is:

  1. Find the differential equation of the first family by eliminating the parameter. For example, x^2 + y^2 = r^2 differentiates to 2x + 2y\,\frac{dy}{dx} = 0, so \frac{dy}{dx} = -\frac{x}{y}.

  2. Find the differential equation of the second family. y = kx gives k = y/x, and differentiating gives \frac{dy}{dx} = k = \frac{y}{x}.

  3. Check that the product of the slopes is -1: \left(-\frac{x}{y}\right)\left(\frac{y}{x}\right) = -1. Confirmed — the families are orthogonal.

Concentric circles and lines through the origin: two orthogonal families. Every line meets every circle at 90°. The radius of a circle is always perpendicular to the tangent — that is the geometric fact driving the orthogonality.

Another important pair. The family of parabolas y^2 = 4ax (one for each a > 0) and the family of parabolas y^2 = 4b(2c - x) can form orthogonal systems under the right conditions. More common in exams is the family of rectangular hyperbolas xy = c^2 (for various c) versus the family x^2 - y^2 = a^2. Differentiating xy = c^2 gives dy/dx = -y/x; differentiating x^2 - y^2 = a^2 gives dy/dx = x/y. The product is (-y/x)(x/y) = -1, so the families are orthogonal.

The rectangular hyperbolas $xy = c$ (black) and the hyperbolas $x^2 - y^2 = a$ (red) form orthogonal families. At every intersection point, the tangent lines are perpendicular.

Common confusions

Going deeper

The core techniques are above — you can stop here for exam preparation. What follows connects these ideas to broader theory.

Envelope of a family of lines

When you write the tangent to y = x^2 at a general point a as y = 2ax - a^2, you have described a one-parameter family of lines, parametrised by a. The parabola itself is the envelope of this family — the curve that every member of the family is tangent to. Envelopes appear across mathematics: the envelope of the normals to a curve is called the evolute, and it is the locus of centres of curvature. For the parabola y = x^2, the evolute turns out to be a cusp-shaped curve. Envelopes will appear again when you study families of circles and conics.

Orthogonal trajectories via differential equations

Finding orthogonal trajectories is one of the first real applications of differential equations. Given a family of curves with differential equation \frac{dy}{dx} = F(x, y), the orthogonal trajectories satisfy \frac{dy}{dx} = -\frac{1}{F(x, y)}. You replace the slope by its negative reciprocal, then solve the resulting ODE. This idea has physical meaning: in electrostatics, the electric field lines and the equipotential surfaces are orthogonal families. In fluid flow, the streamlines and the velocity potential curves are orthogonal. The mathematical structure you learned here — product of slopes equals -1 — is the same structure that organises these physical fields.

When implicit differentiation is essential

For curves given implicitly — say x^3 + y^3 = 3xy (the folium of Descartes) — you cannot write y = f(x) explicitly. But implicit differentiation still gives you \frac{dy}{dx} in terms of x and y. All the techniques above — angle of intersection, tangent from an external point, orthogonality — apply unchanged, with the slope expressed implicitly. The only additional care is that at a self-intersection of the curve, the implicit derivative gives both slopes simultaneously, and you must handle them separately.

Where this leads next