In short

From an external point you can draw exactly two tangents to an ellipse. The line joining their points of contact is the chord of contact, with equation \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 — the same "T = 0" expression. The chord of the ellipse whose midpoint is (h, k) has equation T = S_1. The combined equation of the pair of tangents from an external point is SS_1 = T^2. The reflection property — the tangent at any point makes equal angles with the two focal radii — is why whispering galleries, elliptical pool tables, and medical lithotripters all rely on the ellipse.

A lithotripter is a medical device that breaks kidney stones without surgery. The patient lies inside an elliptical reflector, and the kidney stone is positioned at one focus. A shock wave is generated at the other focus. The wave expands outward, bounces off the elliptical surface, and converges — with devastating precision — at the stone's location. Every ray from one focus, after reflection, passes through the other focus. The machine works because of one geometric fact: the reflection property of the ellipse.

That property was proved in Ellipse — Tangent and Normal. Three more tools follow from the tangent equation — chord of contact, chord with a given midpoint, and the pair of tangents from an external point — each reducing to a single clean formula with a geometric picture that makes it stick.

Throughout, the standard ellipse is \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 with a > b > 0, foci at (\pm c, 0) where c = \sqrt{a^2 - b^2}, and eccentricity e = c/a. The notation S, S_1, and T will be used systematically:

Pair of tangents from an external point

Take a point P(x_1, y_1) outside the ellipse — meaning S_1 > 0. From P, exactly two tangent lines can be drawn to the ellipse. Each tangent touches the curve at one point. The combined equation of both tangent lines is:

Pair of tangents

The combined equation of the two tangent lines from an external point (x_1, y_1) to the ellipse \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 is

\boxed{SS_1 = T^2}

That is:

\left(\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1\right)\left(\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1\right) = \left(\frac{xx_1}{a^2} + \frac{yy_1}{b^2} - 1\right)^2

Derivation. Any line through P(x_1, y_1) can be written as \dfrac{x - x_1}{\cos\alpha} = \dfrac{y - y_1}{\sin\alpha} = r, where r is the distance from P to a generic point (x, y) on the line, and \alpha is the angle the line makes with the x-axis. The parametric point on the line is x = x_1 + r\cos\alpha, y = y_1 + r\sin\alpha.

Substitute into the ellipse equation:

\frac{(x_1 + r\cos\alpha)^2}{a^2} + \frac{(y_1 + r\sin\alpha)^2}{b^2} = 1

Expand:

\frac{x_1^2}{a^2} + \frac{2x_1 r\cos\alpha}{a^2} + \frac{r^2\cos^2\alpha}{a^2} + \frac{y_1^2}{b^2} + \frac{2y_1 r\sin\alpha}{b^2} + \frac{r^2\sin^2\alpha}{b^2} = 1

Group by powers of r:

r^2\left(\frac{\cos^2\alpha}{a^2} + \frac{\sin^2\alpha}{b^2}\right) + 2r\left(\frac{x_1\cos\alpha}{a^2} + \frac{y_1\sin\alpha}{b^2}\right) + \left(\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1\right) = 0

Why: this is a quadratic in r. Its two roots r_1, r_2 are the distances from P to the two intersection points of the line with the ellipse.

For the line to be a tangent, the two intersection points must coincide: r_1 = r_2. This means the discriminant of the quadratic in r must be zero. Setting \Delta = 0:

4\left(\frac{x_1\cos\alpha}{a^2} + \frac{y_1\sin\alpha}{b^2}\right)^2 - 4\left(\frac{\cos^2\alpha}{a^2} + \frac{\sin^2\alpha}{b^2}\right)\left(\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1\right) = 0

This is a condition on \alpha (the direction of the tangent line from P). Now substitute the parametric line x = x_1 + r\cos\alpha, y = y_1 + r\sin\alpha back into the discriminant condition. The \cos\alpha and \sin\alpha terms can be replaced by expressions in (x - x_1) and (y - y_1) divided by r. After this substitution and simplification (expanding and collecting, using the identities \cos^2\alpha + \sin^2\alpha = 1), the r and \alpha both cancel, and you are left with a clean Cartesian equation — the locus of all points on both tangent lines:

\left(\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1\right)\left(\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1\right) = \left(\frac{xx_1}{a^2} + \frac{yy_1}{b^2} - 1\right)^2

The left side is SS_1 and the right side is T^2. So the combined equation is SS_1 = T^2.

What this equation represents. The equation SS_1 = T^2 is a second-degree equation in x and y. It factors (over the reals) into two linear factors — the two tangent lines. Its zero set is the pair of tangent lines from (x_1, y_1). It works for circles, parabolas, and hyperbolas too, with the appropriate S, S_1, and T.

Two tangent lines from the external point $P(7, 3)$ to the ellipse, touching it at $A$ and $B$. The dashed line $AB$ is the chord of contact.

Chord of contact

The two tangent lines from an external point P(x_1, y_1) touch the ellipse at two points — call them A and B. The line segment AB is called the chord of contact of the tangents from P.

Derivation. Let A = (h_1, k_1) and B = (h_2, k_2) be the two points of tangency. The tangent at A is \dfrac{xh_1}{a^2} + \dfrac{yk_1}{b^2} = 1. Since this tangent passes through P(x_1, y_1):

\frac{x_1 h_1}{a^2} + \frac{y_1 k_1}{b^2} = 1 \tag{1}

Similarly, the tangent at B passes through P:

\frac{x_1 h_2}{a^2} + \frac{y_1 k_2}{b^2} = 1 \tag{2}

Both equations (1) and (2) say that the points (h_1, k_1) and (h_2, k_2) satisfy the equation \dfrac{xx_1}{a^2} + \dfrac{yy_1}{b^2} = 1. Since A and B both lie on this line, the line through A and B is:

Chord of contact

The chord of contact of tangents drawn from an external point (x_1, y_1) to the ellipse is

\boxed{T = 0 \;:\; \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1}

Why: the equation looks identical to the tangent at a point on the ellipse, but P(x_1, y_1) is not on the ellipse. When P is on the ellipse, this gives the tangent; when P is outside, it gives the chord of contact. The same formula, two different geometric meanings depending on where P sits.

This is a powerful result. To find the chord of contact, you do not need to find the two tangent points separately and then join them. The single equation T = 0 gives the line directly.

Tangent lines from the external point $P(0, 5)$ to the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$. Since $P$ is on the $y$-axis, symmetry gives two tangent points at equal distances from the $y$-axis. The chord of contact is $T = 0$: $\frac{5y}{9} = 1$, i.e., $y = 9/5 = 1.8$. The chord is horizontal.

Chord with a given midpoint

Sometimes you know the midpoint of a chord and want to find the chord's equation. The result is clean: the chord of the ellipse whose midpoint is (h, k) has the equation T = S_1, where T and S_1 are evaluated at (h, k).

Derivation. Let the chord have endpoints A(x_1, y_1) and B(x_2, y_2), both on the ellipse. Then:

\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1 \qquad \text{and} \qquad \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} = 0

Why: subtracting the two ellipse equations and factoring as differences of squares isolates the slope of the chord.

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

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

So the slope of the chord with midpoint (h, k) is m = -\dfrac{b^2 h}{a^2 k}.

The chord passes through (h, k) with this slope:

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

The left side is T + 1 (where T = \frac{hx}{a^2} + \frac{ky}{b^2} - 1), and the right side is S_1 + 1 (where S_1 = \frac{h^2}{a^2} + \frac{k^2}{b^2} - 1). So T + 1 = S_1 + 1, which gives:

Chord with a given midpoint

The chord of the ellipse whose midpoint is (h, k) has the equation

\boxed{T = S_1}

That is:

\frac{xh}{a^2} + \frac{yk}{b^2} - 1 = \frac{h^2}{a^2} + \frac{k^2}{b^2} - 1

This formula works regardless of whether (h, k) is inside or outside the ellipse (though for (h, k) outside, the "chord" has no real endpoints on the curve, and the equation is that of a line that does not actually intersect the ellipse as a chord).

The reflection property

The tangent and normal article proved that the tangent at any point P on the ellipse makes equal angles with the focal radii PF_1 and PF_2. Here is the consequence for real-world optics: any signal (light, sound, or shock wave) traveling from one focus and striking the ellipse is reflected toward the other focus.

Three different rays from $F_1$, bouncing off the ellipse at $P_1$, $P_2$, and $P_3$, all converge at $F_2$. The reflection property guarantees this for every point on the curve, regardless of where the ray hits.

Applications in India and elsewhere. The whispering gallery at Gol Gumbaz in Bijapur (Karnataka) works on this principle — the dome's cross-section approximates an ellipse, and sound from one focal region reaches the other with startling clarity. Modern lithotripters use the same geometry in reverse: a shock wave is generated at one focus of an ellipsoidal reflector, bounces off the surface, and concentrates at the other focus — where the kidney stone sits. The stone shatters without any incision.

Why the reflection property holds: a physical argument

Imagine stretching a taut string from F_1 to P to F_2, where P is on the ellipse. The string has total length PF_1 + PF_2 = 2a, which is constant. Now let P vary slightly along the ellipse. The total string length stays constant — so the string is at a "stationary" length. By Fermat's principle (or Hero's principle for reflection), a light ray from F_1 to F_2 via a reflecting surface takes the path of stationary optical length. Since every point on the ellipse has the same total focal distance, the ellipse is exactly the surface that reflects F_1-rays to F_2.

The tangent being the reflecting surface means the angle of incidence (measured from the tangent) equals the angle of reflection. This is the reflection property, restated in physical language.

Worked examples

Example 1: Chord of contact from an external point

Find the equation of the chord of contact of tangents drawn from the point P(6, 4) to the ellipse \dfrac{x^2}{25} + \dfrac{y^2}{16} = 1. Also find the length of this chord.

Step 1. Verify that P is outside the ellipse.

S_1 = \frac{36}{25} + \frac{16}{16} - 1 = 1.44 + 1 - 1 = 1.44 > 0 \checkmark

Why: the chord of contact formula applies only when the point is external — that is, S_1 > 0.

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

\frac{6x}{25} + \frac{4y}{16} = 1
\frac{6x}{25} + \frac{y}{4} = 1

Multiply by 100: 24x + 25y = 100.

Why: the formula \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 gives the chord of contact directly, with (x_1, y_1) = (6, 4).

Step 3. Find the endpoints by solving simultaneously with the ellipse. From 24x + 25y = 100, we get y = \dfrac{100 - 24x}{25} = 4 - \dfrac{24x}{25}.

Substitute into the ellipse:

\frac{x^2}{25} + \frac{(4 - 24x/25)^2}{16} = 1
\frac{x^2}{25} + \frac{(100 - 24x)^2}{25^2 \cdot 16} = 1

Multiply by 25 \cdot 16 = 400:

16x^2 + \frac{(100 - 24x)^2}{25} = 400

Multiply by 25:

400x^2 + (100 - 24x)^2 = 10000
400x^2 + 10000 - 4800x + 576x^2 = 10000
976x^2 - 4800x = 0
x(976x - 4800) = 0

So x = 0 or x = \dfrac{4800}{976} = \dfrac{300}{61}.

When x = 0: y = 4. When x = \dfrac{300}{61}: y = 4 - \dfrac{24 \cdot 300}{25 \cdot 61} = 4 - \dfrac{7200}{1525} = 4 - \dfrac{288}{61} = \dfrac{244 - 288}{61} = -\dfrac{44}{61}.

The two contact points are A(0, 4) and B\!\left(\dfrac{300}{61}, -\dfrac{44}{61}\right).

Step 4. Compute the chord length.

AB = \sqrt{\left(\frac{300}{61}\right)^2 + \left(4 + \frac{44}{61}\right)^2} = \sqrt{\frac{300^2}{61^2} + \frac{288^2}{61^2}} = \frac{1}{61}\sqrt{90000 + 82944} = \frac{\sqrt{172944}}{61}

172944 = 16 \times 10809 = 16 \times 9 \times 1201 = 144 \times 1201. So \sqrt{172944} = 12\sqrt{1201}.

AB = \frac{12\sqrt{1201}}{61} \approx \frac{12 \times 34.66}{61} \approx 6.82

Result: The chord of contact is 24x + 25y = 100, with length \dfrac{12\sqrt{1201}}{61}.

The chord of contact $24x + 25y = 100$ (dashed red) of tangents from $P(6, 4)$. The tangent lines from $P$ touch the ellipse at $A(0, 4)$ and $B \approx (4.92, -0.72)$. The chord $AB$ is the dashed line connecting the two contact points.

Notice that A(0, 4) is the top of the minor axis — the tangent there is horizontal. One of the two tangent lines from P(6, 4) to the ellipse is the horizontal line y = 4, which touches the curve at its topmost point. The other tangent has a steeper slope and touches the curve in the fourth quadrant.

Example 2: Chord with a given midpoint

Find the equation of the chord of the ellipse \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1 whose midpoint is (1, 1).

Step 1. Identify a^2 = 9, b^2 = 4, (h, k) = (1, 1).

Why: the midpoint formula T = S_1 uses the midpoint coordinates directly — no need to find the endpoints first.

Step 2. Compute T and S_1 at (h, k) = (1, 1).

T = \frac{x \cdot 1}{9} + \frac{y \cdot 1}{4} - 1 = \frac{x}{9} + \frac{y}{4} - 1
S_1 = \frac{1}{9} + \frac{1}{4} - 1 = \frac{4 + 9 - 36}{36} = -\frac{23}{36}

Step 3. Apply T = S_1.

\frac{x}{9} + \frac{y}{4} - 1 = -\frac{23}{36}
\frac{x}{9} + \frac{y}{4} = 1 - \frac{23}{36} = \frac{13}{36}

Multiply by 36: 4x + 9y = 13.

Step 4. Verify by finding endpoints. Substitute x = \dfrac{13 - 9y}{4} into the ellipse:

\frac{(13 - 9y)^2}{16 \cdot 9} + \frac{y^2}{4} = 1
\frac{(13 - 9y)^2}{144} + \frac{y^2}{4} = 1

Multiply by 144: (13 - 9y)^2 + 36y^2 = 144.

169 - 234y + 81y^2 + 36y^2 = 144

117y^2 - 234y + 25 = 0

The sum of the roots is y_1 + y_2 = \dfrac{234}{117} = 2, so the average is 1. The midpoint has y-coordinate 1. Similarly, x_1 + x_2 = 2 \times 1 = 2. Both match (h, k) = (1, 1).

Result: The chord with midpoint (1, 1) is 4x + 9y = 13.

The chord $4x + 9y = 13$ of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$, with midpoint $M(1, 1)$. The chord has slope $-4/9$, which matches the formula $-\frac{b^2 h}{a^2 k} = -\frac{4 \cdot 1}{9 \cdot 1} = -\frac{4}{9}$.

The slope -4/9 confirms the formula: the chord through (h, k) always has slope -\dfrac{b^2 h}{a^2 k}, which is the same as the slope of the tangent at (h, k) if (h, k) were on the ellipse. The chord is parallel to the tangent at the point with the same x/y ratio.

Common confusions

Going deeper

If you came here for the chord of contact, midpoint formula, and pair of tangents, you have them all. The rest of this section connects these tools to the pole-polar duality and shows how the ellipse's reflection property extends to three dimensions.

Pole and polar

The formula T = 0 defines a line for every point in the plane, whether on the ellipse, outside, or inside. This line is called the polar of the point (the pole) with respect to the ellipse. The relationship is:

Position of the pole The polar is...
On the ellipse The tangent at that point
Outside the ellipse The chord of contact
Inside the ellipse A line outside the ellipse

The pole-polar relationship has a beautiful reciprocity: if the polar of point P passes through point Q, then the polar of Q passes through P. This duality is central to projective geometry and appears in many JEE locus problems.

Pole-polar duality. The polar of the external point $P(6, 3)$ is a chord of contact (dashed red). The polar of the interior point $Q(1.2, 0.6)$ is a line that lies entirely outside the ellipse (dotted). If $P$'s polar passes through $Q$, then $Q$'s polar passes through $P$ — the reciprocity at work.

The reflection property in three dimensions

Rotate the ellipse around its major axis and you get a three-dimensional surface called an ellipsoid of revolution (or prolate spheroid). The reflection property extends: any ray from one focus, after bouncing off the inner surface, passes through the other focus. This is the geometry behind:

The key insight is that the reflection property is a local property of the tangent plane at each point, and the tangent plane to a surface of revolution at any point contains the tangent line to the generating curve. So the two-dimensional property (equal angles with focal radii) lifts directly to three dimensions.

The SS_1 = T^2 identity as an algebraic identity

The equation SS_1 = T^2 looks like it might be a special trick for conics, but it is actually a general algebraic identity. For any second-degree curve S = 0, any point (x_1, y_1) with S_1 = S(x_1, y_1), and T defined by the replacement rule x^2 \to xx_1, y^2 \to yy_1, xy \to \frac{xy_1 + x_1 y}{2}, x \to \frac{x + x_1}{2}, y \to \frac{y + y_1}{2}, the identity SS_1 = T^2 always holds on the pair of tangent lines from (x_1, y_1). This is why the same formula works for circles, parabolas, ellipses, and hyperbolas — the algebraic structure is the same across all conics.

Where this leads next

You now have the complete set of tangent-related tools for the ellipse. The natural next steps: