In short

From any point outside a parabola you can draw exactly two tangents. The line joining their two contact points is the chord of contact; its equation follows directly from the tangent condition. You can also find the chord whose midpoint is a given point, construct the pair of tangents as a single second-degree equation, and use the parabola's reflection property — the fact that every ray parallel to the axis bounces through the focus — which is why satellite dishes and car headlights are parabolic.

A satellite dish is a shallow bowl whose cross-section is a parabola. Every signal arriving parallel to its axis — from a satellite thousands of kilometres away — bounces off the surface and lands at a single point: the focus, where the receiver sits. One curve, one point, perfect concentration.

That is the reflection property of a parabola, and it is one of the results in this article. But before you get there, you need three algebraic tools — chord of contact, chord with a given midpoint, and the pair of tangents from an external point — that appear constantly in JEE problems. Each of them reduces to a single clean formula once you know the parametric form of the parabola.

Throughout this article, the standard parabola is y^2 = 4ax with focus at (a, 0) and directrix x = -a. You should be comfortable with the parametric point (at^2, 2at) and the tangent ty = x + at^2 at that point.

Pair of tangents from an external point

Take the parabola y^2 = 4ax and a point P(x_1, y_1) lying outside it — meaning y_1^2 - 4ax_1 > 0. From P you can draw exactly two tangent lines to the parabola. Each tangent touches the curve at one point. How do you find the combined equation of both tangent lines at once?

The tangent at the parametric point (at^2, 2at) is

ty = x + at^2

If this tangent passes through P(x_1, y_1), then

ty_1 = x_1 + at^2

Rearrange: at^2 - ty_1 + x_1 = 0. This is a quadratic in t. Call its two roots t_1 and t_2 — they correspond to the two points of tangency. By Vieta's formulas:

t_1 + t_2 = \frac{y_1}{a}, \qquad t_1 t_2 = \frac{x_1}{a}

These relations are worth remembering. They connect the external point to the two parameters of the contact points, and they will reappear in every formula below.

The combined equation of the pair of tangents from (x_1, y_1) to y^2 = 4ax is

SS_1 = T^2

where S = y^2 - 4ax, S_1 = y_1^2 - 4ax_1, and T = yy_1 - 2a(x + x_1).

Pair of tangents

The combined equation of the two tangent lines drawn from an external point (x_1, y_1) to the parabola y^2 = 4ax is

\boxed{(y^2 - 4ax)(y_1^2 - 4ax_1) = [yy_1 - 2a(x + x_1)]^2}

This is a second-degree equation in x and y whose zero set is exactly the union of the two tangent lines.

Why this works. Any point (x, y) on a tangent from P satisfies both the line equation and the condition that the line touches the parabola. The expression T = 0 is the chord of contact (derived next), and the identity SS_1 = T^2 factors the pair of tangent lines through the relationship between a point and the conic. You can verify it by expanding both sides and checking that the resulting locus is indeed a pair of lines through (x_1, y_1).

The angle between these two tangents can be computed from the formula

\tan\theta = \frac{t_1 - t_2}{1 + t_1 t_2}

using t_1 - t_2 = \pm\sqrt{(t_1+t_2)^2 - 4t_1 t_2}.

Chord of contact

The two tangent lines from P(x_1, y_1) touch the parabola at two points. The straight line joining those two contact points is called the chord of contact of the tangents from P.

The tangent at (at_1^2, 2at_1) is t_1 y = x + at_1^2, and the tangent at (at_2^2, 2at_2) is t_2 y = x + at_2^2.

The chord joining the contact points (at_1^2, 2at_1) and (at_2^2, 2at_2) has the equation

y(t_1 + t_2) = 2x + 2at_1 t_2

Substitute t_1 + t_2 = y_1/a and t_1 t_2 = x_1/a:

y \cdot \frac{y_1}{a} = 2x + 2a \cdot \frac{x_1}{a}
yy_1 = 2ax + 2x_1 a

Chord of contact

The chord of contact of the tangents drawn from an external point (x_1, y_1) to the parabola y^2 = 4ax is

\boxed{yy_1 = 2a(x + x_1)}

This is the equation T = 0, where T = yy_1 - 2a(x + x_1).

Notice how compact this is. The chord of contact has the same form as the tangent equation at a point on the parabola — you just replace the point on the curve with the external point. This pattern — called the T = 0 rule — works for all conics, not just the parabola.

The parabola $y^2 = 8x$ (so $a = 2$). From the external point $P(4, 6)$, two tangent lines touch the curve at $A(2, 4)$ and $B(8, 8)$. The dashed line $AB$ is the chord of contact, with equation $6y = 4(x + 4)$, or $3y = 2x + 8$. The tangent at $A$ (parameter $t = 1$) is $y = x + 2$; the tangent at $B$ (parameter $t = 2$) is $2y = x + 8$.

Chord with a given midpoint

A different question: given a point M(x_1, y_1) inside the parabola, find the equation of the chord of y^2 = 4ax whose midpoint is M.

Let the endpoints of the chord be (at_1^2, 2at_1) and (at_2^2, 2at_2). The midpoint conditions are:

\frac{at_1^2 + at_2^2}{2} = x_1, \qquad \frac{2at_1 + 2at_2}{2} = y_1

From the second equation: t_1 + t_2 = y_1/a.

The slope of the chord is

m = \frac{2at_1 - 2at_2}{at_1^2 - at_2^2} = \frac{2a(t_1 - t_2)}{a(t_1 - t_2)(t_1 + t_2)} = \frac{2}{t_1 + t_2} = \frac{2a}{y_1}

Why: the factor (t_1 - t_2) cancels from numerator and denominator, leaving a slope that depends only on the sum t_1 + t_2, which equals y_1/a.

The chord passes through (x_1, y_1) with slope 2a/y_1:

y - y_1 = \frac{2a}{y_1}(x - x_1)

Multiply through by y_1:

y_1(y - y_1) = 2a(x - x_1)
yy_1 - y_1^2 = 2ax - 2ax_1

Chord with a given midpoint

The equation of the chord of the parabola y^2 = 4ax whose midpoint is (x_1, y_1) is

\boxed{yy_1 - 2a(x + x_1) = y_1^2 - 4ax_1}

In shorthand: T = S_1, where T = yy_1 - 2a(x + x_1) and S_1 = y_1^2 - 4ax_1.

The three formulas — T = 0 for chord of contact, T = S_1 for chord with given midpoint, SS_1 = T^2 for pair of tangents — form a clean trio. They use the same building blocks S, S_1, and T, just assembled differently. This is one of the deepest structural patterns in conic section geometry, and it carries over unchanged to the ellipse and hyperbola.

The chord of $y^2 = 8x$ whose midpoint is $M(3, 2)$. Using $T = S_1$: $2y - 4(x + 3) = 4 - 24$, giving $2y = 4x - 8$, or $y = 2x - 4$. The chord endpoints, found by solving $y = 2x - 4$ with $y^2 = 8x$, have $x$-coordinates $3 \pm \sqrt{5}$, and the midpoint is indeed $(3, 2)$.

Worked examples

Example 1: Chord of contact from a given point

Find the equation of the chord of contact of the tangents drawn from the point P(9, 12) to the parabola y^2 = 12x.

Step 1. Identify a and check that P is external. Compare y^2 = 12x with y^2 = 4ax: 4a = 12, so a = 3. Check: S_1 = 144 - 12(9) = 144 - 108 = 36 > 0. The point is outside the parabola.

Why: every formula uses a, so extract it first. The sign of S_1 = y_1^2 - 4ax_1 tells you whether the point is outside (S_1 > 0), on (S_1 = 0), or inside (S_1 < 0) the parabola.

Step 2. Write the chord of contact equation T = 0: yy_1 = 2a(x + x_1).

12y = 6(x + 9) \implies 2y = x + 9

Why: substitute (x_1, y_1) = (9, 12) and a = 3 directly into the formula, then simplify.

Step 3. Find the contact points. The tangent at parameter t is ty = x + 3t^2. If it passes through (9, 12):

12t = 9 + 3t^2 \implies t^2 - 4t + 3 = 0 \implies (t-1)(t-3) = 0

So t_1 = 1 and t_2 = 3. The contact points are A(3, 6) and B(27, 18).

Why: each root t gives a parametric point (at^2, 2at) = (3t^2, 6t). At t = 1: (3, 6). At t = 3: (27, 18).

Step 4. Verify that both contact points lie on the chord 2y = x + 9. At A(3, 6): 12 = 12. At B(27, 18): 36 = 36. Both check out.

Result: The chord of contact from (9, 12) to y^2 = 12x is 2y = x + 9.

From $P(9, 12)$, two tangents touch $y^2 = 12x$ at $A(3, 6)$ and $B(27, 18)$. The dashed chord of contact $AB$ has equation $2y = x + 9$. Both tangent lines and the chord form a triangle — a figure that appears constantly in JEE problems.

The triangle PAB formed by the two tangent lines and the chord of contact is a recurring configuration in conic problems. Its area can be computed directly from the coordinates.

Example 2: Pair of tangents from an external point

Find the combined equation of the pair of tangents drawn from P(1, 4) to the parabola y^2 = 4x.

Step 1. Identify a. Here 4a = 4, so a = 1. Verify that P is external: S_1 = 16 - 4(1) = 12 > 0. Good.

Why: if S_1 \leq 0, the point is on or inside the parabola, and the pair-of-tangents formula does not apply.

Step 2. Compute S, S_1, and T.

S = y^2 - 4x, \qquad S_1 = 16 - 4 = 12, \qquad T = 4y - 2(x + 1) = 4y - 2x - 2

Why: T = yy_1 - 2a(x + x_1) with (x_1, y_1) = (1, 4) and a = 1.

Step 3. Apply SS_1 = T^2.

12(y^2 - 4x) = (4y - 2x - 2)^2

Expand the right side: (4y - 2x - 2)^2 = 16y^2 - 16xy - 16y + 4x^2 + 8x + 4.

So: 12y^2 - 48x = 16y^2 - 16xy - 16y + 4x^2 + 8x + 4.

Why: expand carefully — every cross term matters.

Step 4. Rearrange to standard form.

0 = 4x^2 + 4y^2 - 16xy - 16y + 56x + 4

Divide by 4:

x^2 + y^2 - 4xy - 4y + 14x + 1 = 0

This can be factored (with effort) into the product of two linear equations, each representing one tangent line. For verification: the tangent at parameter t passing through (1, 4) gives t^2 - 4t + 1 = 0, so t = 2 \pm \sqrt{3}. The two tangent lines are y(2+\sqrt{3}) = x + (7 + 4\sqrt{3}) and y(2-\sqrt{3}) = x + (7 - 4\sqrt{3}), and their combined equation matches.

Result: The pair of tangents from (1, 4) to y^2 = 4x is x^2 + y^2 - 4xy - 4y + 14x + 1 = 0.

The parabola $y^2 = 4x$ with the two tangent lines from $P(1, 4)$. The steeper tangent corresponds to $t = 2 + \sqrt{3}$ and the flatter one to $t = 2 - \sqrt{3}$. Both lines pass through $P$ and touch the curve at exactly one point each.

The combined equation SS_1 = T^2 encodes both lines in a single second-degree equation. Factoring it back into two separate lines is possible but rarely necessary in practice — most problems only need the combined form.

The reflection property

This is the result that explains satellite dishes, car headlights, solar concentrators, and the whispering galleries of old forts.

Claim. Any ray travelling parallel to the axis of a parabola, after reflecting off the curve, passes through the focus.

The proof uses the tangent and normal at a point on the parabola.

Take a point P(at^2, 2at) on y^2 = 4ax. The tangent at P has slope 1/t (from differentiating y^2 = 4ax implicitly, or from the parametric tangent equation). The normal at P therefore has slope -t.

A ray arrives parallel to the axis — that is, horizontally, in the direction of increasing x. When it hits the curve at P, it reflects off the surface. By the law of reflection, the angle of incidence equals the angle of reflection, both measured from the tangent (or equivalently, from the normal).

The claim is that the reflected ray passes through the focus F(a, 0). Here is a clean proof using angles with the tangent.

The slope of the line from P(at^2, 2at) to the focus F(a, 0) is

m_{PF} = \frac{2at - 0}{at^2 - a} = \frac{2t}{t^2 - 1}

The tangent at P has slope 1/t. Compute the angle the line PF makes with the tangent:

\tan\phi_1 = \frac{m_{PF} - 1/t}{1 + m_{PF}/t} = \frac{\frac{2t}{t^2-1} - \frac{1}{t}}{1 + \frac{2}{t^2-1}} = \frac{\frac{2t^2 - t^2 + 1}{t(t^2-1)}}{\frac{t^2 - 1 + 2}{t^2-1}} = \frac{\frac{t^2+1}{t(t^2-1)}}{\frac{t^2+1}{t^2-1}} = \frac{1}{t}

The incoming horizontal ray has slope 0. Its angle with the tangent (slope 1/t) is

\tan\phi_2 = \frac{0 - 1/t}{1 + 0} = -\frac{1}{t}

So |\tan\phi_1| = |\tan\phi_2| = 1/t. The line PF and the incoming horizontal ray make equal angles with the tangent at P. By the law of reflection, the reflected ray is exactly PF. Every horizontal ray bounces through the focus.

The reflection property of $y^2 = 8x$ (focus at $F(2, 0)$). Three horizontal rays (dashed) hit the parabola at $P$, $Q$, and $R$. Each reflected ray (red) passes through the focus $F$. This is why a parabolic dish concentrates all incoming parallel signals at a single point.

The converse is equally useful. If you place a light source at the focus, every ray hits the parabola and reflects outward parallel to the axis — producing a perfectly collimated beam. This is the principle behind car headlights, searchlights, and the giant parabolic mirrors used in solar thermal power plants in Rajasthan's Thar Desert.

Why only the parabola has this property

Among all conic sections, the parabola is the only curve with a single focus whose reflection property involves parallel rays. An ellipse has two foci, and its reflection property sends rays from one focus to the other (this is what makes whispering galleries work). A hyperbola also has two foci, and incoming rays aimed at one focus reflect toward the other. The parabola is the limiting case — a conic with eccentricity exactly 1 — where the second focus has moved to infinity. "Parallel rays" are rays aimed at a point at infinity.

A structural note: S, S_1, T

The three key results of this article share a common language. Define, for the parabola y^2 = 4ax:

Then:

Problem Formula
Chord of contact from (x_1, y_1) T = 0
Chord with midpoint (x_1, y_1) T = S_1
Pair of tangents from (x_1, y_1) SS_1 = T^2
Tangent at (x_1, y_1) on the curve T = 0 (same formula, but S_1 = 0 since the point is on the curve)

This trio of formulas works identically for the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 and the hyperbola \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 — only the definitions of S, S_1, and T change. Recognising the pattern once means you never have to re-derive these results for a different conic.

The same parabola $y^2 = 8x$ with three types of distinguished points. $P(4, 6)$ is external ($S_1 > 0$): use $T = 0$ for chord of contact, $SS_1 = T^2$ for pair of tangents. $M(3, 2)$ is interior ($S_1 < 0$): use $T = S_1$ for the chord with midpoint $M$. $Q(2, 4)$ is on the curve ($S_1 = 0$): $T = 0$ gives the tangent at $Q$. The formula you use depends on where the point sits relative to the parabola.

Common confusions

Going deeper

If you are comfortable with the chord of contact, the midpoint chord, the pair of tangents, and the reflection property, you have covered the main results. The rest of this section connects these ideas to locus problems and a classical length formula.

Locus of the midpoint of a focal chord

A focal chord is any chord that passes through the focus. For y^2 = 4ax, let a chord through F(a, 0) have endpoints (at_1^2, 2at_1) and (at_2^2, 2at_2). The condition for the chord to pass through the focus gives t_1 t_2 = -1 (this is a standard result from the tangent-and-normal article).

The midpoint (h, k) of the focal chord satisfies:

h = \frac{a(t_1^2 + t_2^2)}{2}, \qquad k = a(t_1 + t_2)

From t_1 t_2 = -1: t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1 t_2 = (t_1 + t_2)^2 + 2.

Let s = t_1 + t_2. Then h = a(s^2 + 2)/2 and k = as, so s = k/a.

h = \frac{a(k^2/a^2 + 2)}{2} = \frac{k^2}{2a} + a
k^2 = 2a(h - a)

Replacing (h, k) with (x, y):

y^2 = 2a(x - a)

This is itself a parabola — with the same axis as the original, but shifted right by a and with half the "width" parameter.

Length of the chord of contact

If the tangents from an external point P(x_1, y_1) touch y^2 = 4ax at parameters t_1 and t_2, the length of the chord of contact is

\ell = a|t_1 - t_2|\sqrt{(t_1 + t_2)^2 + 4}

Using t_1 + t_2 = y_1/a and t_1 t_2 = x_1/a:

|t_1 - t_2|^2 = (t_1 + t_2)^2 - 4t_1 t_2 = \frac{y_1^2}{a^2} - \frac{4x_1}{a} = \frac{y_1^2 - 4ax_1}{a^2} = \frac{S_1}{a^2}

So:

\ell = \frac{\sqrt{S_1}}{a}\sqrt{\frac{y_1^2}{a^2} + 4} = \frac{1}{a}\sqrt{S_1}\sqrt{y_1^2 + 4a^2}/a = \frac{\sqrt{S_1(y_1^2 + 4a^2)}}{a^2}

This formula lets you compute the chord length directly from the coordinates of the external point, without finding the contact points explicitly.

The optical property in Indian architecture

The parabolic reflection property was understood empirically long before it was proved mathematically. The curved surfaces of the Jantar Mantar instruments in Jaipur and Delhi, built by Maharaja Jai Singh II in the 1720s, use precisely computed curves to focus sunlight for accurate time measurement. While these instruments use various conic sections, the underlying principle — that curved surfaces can direct light in predictable ways — is the same one at work in this article.

Where this leads next

The chord of contact, midpoint chord, and pair-of-tangent formulas you learned here are not specific to the parabola — the same structural pattern (T = 0, T = S_1, SS_1 = T^2) applies to all conics. The next set of articles extends these ideas to curves with two foci.