In short

From a point outside a hyperbola, exactly two tangent lines can be drawn to the curve. The chord joining those two contact points is called the chord of contact, and its equation is T = 0 — the same substitution trick used for tangents. The equation of the chord whose midpoint is a given point uses the T = S_1 form. And the pair of tangents from an external point satisfies SS_1 = T^2.

Imagine standing on a hilltop and looking at two straight roads that stretch into the distance. Each road just barely touches a curved lake at one point before veering away. Someone asks: what is the shortest path from one touching point to the other, staying over the lake? That path — the chord connecting the two tangent points — is the chord of contact, and it shows up everywhere in conic section problems.

Stand at the point (6, 3) and look at the hyperbola \dfrac{x^2}{9} - \dfrac{y^2}{4} = 1. You are outside the curve — you can verify: \dfrac{36}{9} - \dfrac{9}{4} = 4 - 2.25 = 1.75 > 1, which means the point lies in the exterior. From where you stand, you can draw exactly two lines that just graze the curve without cutting through it. Those are the two tangent lines from your point.

Now connect the two points where those tangent lines touch the hyperbola. That connecting line is the chord of contact. And there is something remarkable about it: its equation has exactly the same form as the tangent equation, using the external point's coordinates. The pattern that governs tangent lines — the T = 0 substitution — governs the chord of contact too.

This article covers four constructions that extend the tangent theory: chord of contact, chord with a given midpoint, the pair of tangents from an external point, and the reflection property. Each one follows from the same algebraic machinery, and each one appears regularly in JEE and competitive problems. Underlying all of them is a single unifying idea — the pole-polar correspondence — which ties the whole theory together.

The notation

Before the derivations, set up the shorthand that makes the algebra readable. For the hyperbola S \equiv \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} - 1 = 0:

If (x_1, y_1) is on the hyperbola, S_1 = 0 and T = 0 is the tangent. If (x_1, y_1) is outside the hyperbola, S_1 > 0 and T = 0 is the chord of contact. If (x_1, y_1) is inside the hyperbola, S_1 < 0.

This notation is shared across all conics — circles, parabolas, ellipses, and hyperbolas all use S, S_1, and T in the same way.

Chord of contact

Suppose you are at an external point (x_1, y_1) — meaning S_1 > 0. You draw the two tangents from this point, touching the hyperbola at points A and B. The line AB is the chord of contact.

Derivation. Let A = (\alpha_1, \beta_1) and B = (\alpha_2, \beta_2) be the two contact points, both on the hyperbola. The tangent at A is:

\frac{x \alpha_1}{a^2} - \frac{y \beta_1}{b^2} = 1

Since this tangent passes through (x_1, y_1):

\frac{x_1 \alpha_1}{a^2} - \frac{y_1 \beta_1}{b^2} = 1 \quad \cdots (1)

Similarly, the tangent at B passes through (x_1, y_1):

\frac{x_1 \alpha_2}{a^2} - \frac{y_1 \beta_2}{b^2} = 1 \quad \cdots (2)

Equations (1) and (2) say that both (\alpha_1, \beta_1) and (\alpha_2, \beta_2) satisfy the equation:

\frac{x_1 X}{a^2} - \frac{y_1 Y}{b^2} = 1

where X and Y are the running coordinates. But the line through two points is unique. So the chord of contact AB has the equation:

\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1

Chord of contact

The chord of contact of tangents drawn from an external point (x_1, y_1) to the hyperbola \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 is:

T = 0 \quad \text{i.e.,} \quad \frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1

This is the same equation as the tangent at a point on the curve — the only difference is that (x_1, y_1) is now outside the curve, not on it.

The elegance here is that the same formula does double duty. When the point is on the curve, the formula gives the tangent. When the point is off the curve, the same formula gives the chord of contact. The algebra does not care — the equation T = 0 adapts to the context.

From the external point $P(6, 3)$, two tangent lines reach the hyperbola (dashed). The chord of contact (red, solid) joins the two tangent contact points. Its equation is given by $T = 0$ using $P$'s coordinates.

Chord with a given midpoint

Here is a different question: you are told that a chord of the hyperbola has its midpoint at the point (h, k). What is the equation of that chord?

Derivation. Let the endpoints of the chord be (x_1, y_1) and (x_2, y_2), both on the hyperbola. Then:

\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1 \quad \text{and} \quad \frac{x_2^2}{a^2} - \frac{y_2^2}{b^2} = 1

Subtract:

\frac{x_1^2 - x_2^2}{a^2} - \frac{y_1^2 - y_2^2}{b^2} = 0
\frac{(x_1 + x_2)(x_1 - x_2)}{a^2} = \frac{(y_1 + y_2)(y_1 - y_2)}{b^2}

The midpoint is (h, k), so x_1 + x_2 = 2h and y_1 + y_2 = 2k. The slope of the chord is m = \dfrac{y_1 - y_2}{x_1 - x_2}. Substituting:

\frac{2h}{a^2} = \frac{2k}{b^2} \cdot m
m = \frac{b^2 h}{a^2 k}

The chord passes through (h, k) with slope \dfrac{b^2 h}{a^2 k}:

y - k = \frac{b^2 h}{a^2 k}(x - h)

Multiply by a^2 k:

a^2 k(y - k) = b^2 h(x - h)
b^2 hx - a^2 ky = b^2 h^2 - a^2 k^2

Divide by a^2 b^2:

\frac{hx}{a^2} - \frac{ky}{b^2} = \frac{h^2}{a^2} - \frac{k^2}{b^2}

The left side is T (with (h, k) playing the role of (x_1, y_1), minus the -1). The right side is S_1 (plus 1). More precisely:

\frac{hx}{a^2} - \frac{ky}{b^2} - 1 = \frac{h^2}{a^2} - \frac{k^2}{b^2} - 1
T = S_1

Chord with a given midpoint

The equation of the chord of the hyperbola \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 whose midpoint is (h, k) is:

T = S_1

i.e., \quad \dfrac{xh}{a^2} - \dfrac{yk}{b^2} - 1 = \dfrac{h^2}{a^2} - \dfrac{k^2}{b^2} - 1

The formula T = S_1 is universal across conics. For any conic, the chord whose midpoint is a given point has the equation T = S_1, where T and S_1 are computed from the conic's equation.

Pair of tangents from an external point

From an external point (x_1, y_1), two tangent lines go to the hyperbola. The combined equation of both tangent lines together — as a single second-degree equation — has a beautiful form.

Derivation. Consider a general point (x, y) in the plane. Draw the line from (x_1, y_1) through (x, y). A general point on this line can be parametrised as:

(x_1 + t(x - x_1),\, y_1 + t(y - y_1))

where t = 0 gives (x_1, y_1) and t = 1 gives (x, y). Substitute this into the hyperbola equation S = 0:

\frac{(x_1 + t(x - x_1))^2}{a^2} - \frac{(y_1 + t(y - y_1))^2}{b^2} - 1 = 0

Expand each squared term. For the x-part:

(x_1 + t(x - x_1))^2 = x_1^2 + 2x_1 t(x - x_1) + t^2(x - x_1)^2

Similarly for the y-part. Collecting powers of t:

t^2 \cdot S_{(x,y)} + 2t \cdot (T_{(x,y)} - S_1) + S_1 \cdot 1 = 0

where S_{(x,y)} denotes S evaluated at (x, y) — which we simply call S — and T_{(x,y)} is T evaluated using (x_1, y_1) as the fixed point. More precisely, after careful expansion, the quadratic in t becomes:

S \cdot t^2 + 2(T - S_1) \cdot t + S_1 = 0

This quadratic has two roots in t, corresponding to the two points where the line from (x_1, y_1) through (x, y) meets the hyperbola. For the line to be tangent, these two intersection points must coincide, so the discriminant of this quadratic in t must be zero:

4(T - S_1)^2 - 4 S \cdot S_1 = 0
(T - S_1)^2 = S \cdot S_1
T^2 - 2T \cdot S_1 + S_1^2 = S \cdot S_1

Now, the locus condition simplifies. Since (x_1, y_1) is the external point (a fixed point, not a variable), S_1 is a constant. The condition that a point (x, y) lies on one of the tangent lines from (x_1, y_1) is:

S \cdot S_1 = T^2

Pair of tangents from an external point

The combined equation of the pair of tangents drawn from an external point (x_1, y_1) to the hyperbola \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 is:

SS_1 = T^2

where S = \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} - 1, S_1 = \dfrac{x_1^2}{a^2} - \dfrac{y_1^2}{b^2} - 1, and T = \dfrac{xx_1}{a^2} - \dfrac{yy_1}{b^2} - 1.

This equation, SS_1 = T^2, is a second-degree equation in x and y. It represents two straight lines (the pair of tangents) taken together. The equation T^2 = SS_1 is a factored form — in principle, you can factor the left minus the right into two linear factors, and those are the individual tangent lines.

The formula works for all conics. For a circle x^2 + y^2 = r^2, for a parabola y^2 = 4ax, for an ellipse — the same SS_1 = T^2.

The pair of tangents from an external point $P(5, 5)$ to the hyperbola $\frac{x^2}{4} - \frac{y^2}{3} = 1$. Both tangent lines touch the same branch. The combined equation of these two lines is $SS_1 = T^2$.

The reflection property

This is the property that makes hyperbolas useful in physics and engineering.

Statement. If P is any point on the hyperbola and S, S' are the two foci, then the tangent at P bisects the exterior angle between the focal radii PS and PS'. Equivalently, the normal at P bisects the interior angle \angle SPS'.

What this means physically. A ray of light travelling toward one focus of a hyperbolic mirror will be reflected so that it appears to come from the other focus.

Proof. Take the hyperbola \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 with foci S(ae, 0) and S'(-ae, 0). Let P = (x_1, y_1) be on the hyperbola. The tangent at P has the equation \dfrac{xx_1}{a^2} - \dfrac{yy_1}{b^2} = 1.

The tangent meets the x-axis at T = (a^2/x_1, 0). Now compute the ratio in which T divides SS' externally.

\frac{ST}{TS'} = \frac{a^2/x_1 - ae}{a^2/x_1 + ae} = \frac{a^2 - aex_1}{a^2 + aex_1} = \frac{a - ex_1}{a + ex_1}

From the focal distance property of the hyperbola, SP = |ex_1 - a| and S'P = ex_1 + a (taking P on the right branch, so x_1 > 0 and ex_1 > a). Therefore:

\frac{ST}{TS'} = \frac{a - ex_1}{a + ex_1} = -\frac{ex_1 - a}{ex_1 + a} = -\frac{SP}{S'P}

The negative sign means T divides SS' externally in the ratio SP : S'P. By the angle bisector theorem (external division), the line PT — the tangent — is the external bisector of \angle SPS'. Therefore, the normal at P is the internal bisector.

This is the same result that holds for ellipses (where the tangent bisects the exterior angle and the normal bisects the interior angle), with one crucial difference: for an ellipse, a ray heading toward one focus reflects to the other focus. For a hyperbola, a ray heading toward one focus reflects away from the other focus — but along the line through the other focus.

Cassegrain telescopes exploit this. A parabolic primary mirror focuses starlight toward its focus. A small hyperbolic secondary mirror, placed so that one of its foci coincides with the parabolic focus, reflects the converging light back through a hole in the primary, toward the hyperbola's other focus where the eyepiece sits.

The reflection property of a hyperbolaA schematic showing a point P on the right branch of a hyperbola, with lines drawn from P to the two foci S and S prime. The tangent line at P bisects the exterior angle between PS and PS prime. The normal at P bisects the interior angle. S' S P tangent normal
The reflection property: the tangent at $P$ bisects the exterior angle between $PS$ and $PS'$. A ray aimed at $S$ reflects off the hyperbola and appears to come from $S'$. The normal bisects the interior angle.

This property is not just a theoretical curiosity. ISRO's satellite communication systems and radio telescope designs use reflectors shaped as conic sections — parabolas for the primary dish and hyperbolas for the secondary reflector — precisely because of these angle-bisection properties.

The interplay between the three formulas

Before moving to examples, pause and notice how the three results — T = 0 for chord of contact, T = S_1 for midpoint chord, and SS_1 = T^2 for pair of tangents — are three faces of the same algebraic coin. All three come from substituting a line into the conic equation to get a quadratic, and then imposing a condition on that quadratic:

In each case, the substitution x^2 \to xx_1, y^2 \to yy_1 (the T operation) appears because it naturally arises when you take the difference of the conic equation at two points, or when you impose the double-root condition. The algebraic structure is the same — only the geometric question changes.

Worked examples

Example 1: Chord of contact from an external point

Find the chord of contact of the tangents drawn from (4, 2) to the hyperbola \dfrac{x^2}{9} - \dfrac{y^2}{4} = 1.

Step 1. Verify the point is external to the hyperbola.

S_1 = \frac{16}{9} - \frac{4}{4} - 1 = \frac{16}{9} - 2 = \frac{-2}{9}

This is negative, meaning (4, 2) is actually inside the curve (in the region between the branches where S < 0). A chord of contact cannot be drawn from an interior point.

Let us use (6, 3) instead.

S_1 = \frac{36}{9} - \frac{9}{4} - 1 = 4 - 2.25 - 1 = 0.75 > 0 \checkmark

Why: the sign of S_1 tells you where the point sits relative to the hyperbola. Positive means exterior (tangents exist); negative means interior (between the branches); zero means on the curve.

Step 2. Write the chord of contact using T = 0.

\frac{6x}{9} - \frac{3y}{4} = 1
\frac{2x}{3} - \frac{3y}{4} = 1

Multiply through by 12: 8x - 9y = 12.

Why: the T = 0 formula directly gives the chord of contact — no need to find the individual tangent lines first.

Step 3. Find where this chord meets the hyperbola.

From 8x - 9y = 12: y = \dfrac{8x - 12}{9}.

Substitute into \dfrac{x^2}{9} - \dfrac{y^2}{4} = 1:

\frac{x^2}{9} - \frac{(8x - 12)^2}{324} = 1
\frac{36x^2 - (8x - 12)^2}{324} = 1
36x^2 - 64x^2 + 192x - 144 = 324
-28x^2 + 192x - 468 = 0
7x^2 - 48x + 117 = 0
x = \frac{48 \pm \sqrt{2304 - 3276}}{14} = \frac{48 \pm \sqrt{-972}}{14}

The discriminant is negative — the chord does not intersect the same branch in real coordinates. This happens when the external point sees one tangent going to each branch, so the chord of contact connects points on different branches.

Step 4. The chord of contact still has a geometric meaning: it is the polar line of the point (6, 3) with respect to the hyperbola, and the two tangent points lie on opposite branches.

Result: The chord of contact is 8x - 9y = 12.

The hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ with the chord of contact (red line) from the external point $P(6, 3)$. The dashed lines suggest the tangent directions from $P$ to the curve.

The chord of contact is the red line. Even though the tangent points may lie on opposite branches, the chord of contact equation T = 0 captures both.

Example 2: Pair of tangents and chord with given midpoint

For the hyperbola \dfrac{x^2}{4} - \dfrac{y^2}{3} = 1, find the equation of the chord whose midpoint is (3, 2).

Step 1. Compute S_1 at the midpoint (3, 2).

S_1 = \frac{9}{4} - \frac{4}{3} - 1 = \frac{27 - 16 - 12}{12} = -\frac{1}{12}

Why: S_1 is needed for the T = S_1 formula. A negative S_1 means the midpoint lies between the branches — and a chord with this midpoint is possible only if both endpoints are on the same branch and the chord passes through the inter-branch region.

Step 2. Compute T (with (h, k) = (3, 2)).

T = \frac{3x}{4} - \frac{2y}{3} - 1

Step 3. Apply T = S_1.

\frac{3x}{4} - \frac{2y}{3} - 1 = -\frac{1}{12}
\frac{3x}{4} - \frac{2y}{3} = 1 - \frac{1}{12} = \frac{11}{12}

Multiply through by 12: 9x - 8y = 11.

Why: the T = S_1 formula gives the chord equation directly from the midpoint coordinates — no need to find the endpoints.

Step 4. Verify the slope. The slope of the chord is 9/8. From the midpoint slope formula: \dfrac{b^2 h}{a^2 k} = \dfrac{3 \times 3}{4 \times 2} = \dfrac{9}{8}. This matches.

Result: The chord with midpoint (3, 2) is 9x - 8y = 11.

The hyperbola $\frac{x^2}{4} - \frac{y^2}{3} = 1$ with the chord (red) whose midpoint is $M(3, 2)$. The chord cuts the right branch at two points equidistant from $M$.

The chord passes through (3, 2) at the slope 9/8 predicted by the formula. Both intersection points with the right branch are symmetric about M, confirming that M is the midpoint.

Common confusions

Going deeper

If you came here to learn chord of contact, midpoint chord, and pair of tangents, you have everything you need for problems. What follows is for readers who want to understand the unifying concept behind all of these — the pole-polar relationship.

The pole and polar — the unifying idea

All four constructions in this article — tangent, chord of contact, chord with midpoint, pair of tangents — are shadows of a single deeper concept: the pole-polar correspondence.

Given a conic S = 0 and a point P = (x_1, y_1) (called the pole), the polar of P with respect to the conic is the line T = 0:

\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1

This is the same equation in all cases. What changes is the position of P:

Position of P S_1 What T = 0 represents
On the conic = 0 Tangent at P
Outside the conic > 0 Chord of contact from P
Inside the conic < 0 Polar line (no direct tangent interpretation)

The pole-polar relationship has a remarkable property called reciprocity: if the polar of P passes through Q, then the polar of Q passes through P. To see why, suppose P = (x_1, y_1) and Q = (x_2, y_2). The polar of P is:

\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1

If Q lies on this line, then:

\frac{x_2 x_1}{a^2} - \frac{y_2 y_1}{b^2} = 1

But this expression is symmetric in the roles of (x_1, y_1) and (x_2, y_2) — it is the same as saying that P lies on the polar of Q. The symmetry of the bilinear expression T is doing all the work.

This reciprocity is the foundation of projective geometry and duality in conics. It means that the relationship between a point and its polar line is two-way — there is no preferred direction.

The director circle

The locus of all external points from which the two tangents to the hyperbola are perpendicular to each other is a circle called the director circle. For the hyperbola \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1, the director circle is:

x^2 + y^2 = a^2 - b^2

This circle exists only when a > b. When a = b (a rectangular hyperbola), the director circle degenerates to a point at the origin. When a < b, there is no real locus — no external point gives perpendicular tangents.

Derivation. The two tangents from (x_1, y_1) in slope form are y_1 = mx_1 \pm \sqrt{a^2 m^2 - b^2}. Rearranging and squaring: (y_1 - mx_1)^2 = a^2 m^2 - b^2. This gives a quadratic in m:

m^2(x_1^2 - a^2) - 2mx_1 y_1 + (y_1^2 + b^2) = 0

The two roots m_1, m_2 are the slopes of the two tangents. For perpendicular tangents, m_1 m_2 = -1:

\frac{y_1^2 + b^2}{x_1^2 - a^2} = -1
y_1^2 + b^2 = -x_1^2 + a^2
x_1^2 + y_1^2 = a^2 - b^2
The hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ with its director circle $x^2 + y^2 = 5$ (dashed red). From any point on the director circle, the two tangents to the hyperbola are perpendicular. Since $a^2 - b^2 = 9 - 4 = 5 > 0$, the director circle exists and has radius $\sqrt{5}$.

For the ellipse, the corresponding result is x^2 + y^2 = a^2 + b^2 — again, the sign difference between ellipse and hyperbola.

The number of tangents from a point

The sign of S_1 determines how many tangent lines can be drawn from (x_1, y_1) to the hyperbola:

This is identical to the situation for ellipses and circles, and the same S_1 test works for all conics.

Comparing across conics — a summary table

Here is the complete comparison of the advanced formulas across all three central conics:

Formula Circle x^2 + y^2 = r^2 Ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 Hyperbola \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
Tangent at point xx_1 + yy_1 = r^2 \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1
Chord of contact Same as tangent formula Same as tangent formula Same as tangent formula
Midpoint chord T = S_1 T = S_1 T = S_1
Pair of tangents SS_1 = T^2 SS_1 = T^2 SS_1 = T^2
Director circle x^2+y^2=2r^2 x^2+y^2=a^2+b^2 x^2+y^2=a^2-b^2

The first four rows are identical in structure — the only thing that changes is the definition of S, S_1, and T for each conic. The director circle is the one place where the sign difference between ellipse and hyperbola produces a qualitatively different result (the hyperbola's may not exist as a real circle).

This universality is why the S-T-S_1 notation is so powerful. Learn it once, and it works for every conic you will ever meet.

Where this leads next

You now have the full advanced toolkit for hyperbolas. The natural continuations: