Parabola latus rectum

Draw a parabola and mark its focus — the special interior point that gives the curve its reflective, "focusing" property. Now draw the chord through that focus that runs perpendicular to the axis. That chord is the latus rectum, and its length is one of the simplest numbers attached to a parabola: it tells you, at a glance, how wide or narrow the curve is right where it matters most, near the focus. This is a short but high-yield corner of the Conic Sections chapter in CBSE Class 11 Mathematics (Unit III, Coordinate Geometry). The result is almost embarrassingly clean — for the standard parabola $y^2 = 4ax$ , the latus rectum has length exactly $4a$ , the same number that already sits on the right-hand side of the equation. This article rebuilds that result from scratch, extends it to all four standard orientations, and works two exam-style problems: reading the length off an equation, and reconstructing the equation from a given length.

The core insight

The latus rectum of a parabola is the focal chord perpendicular to the axis, and for every standard orientation its length equals $4a$ , the coefficient already present in the parabola's standard equation.

The core idea

What the latus rectum is. For a parabola, the latus rectum is defined as the chord that (a) passes through the focus and (b) is perpendicular to the axis of the parabola. Because a parabola is symmetric about its axis, this chord is automatically bisected by the axis — its two endpoints sit at equal distances above and below (or left and right of) the focus.

Deriving the length for $y^2 = 4ax$ . The standard parabola $y^2 = 4ax$ (with $a > 0$ ) has vertex at the origin and focus $S(a, 0)$ . The axis is the $x$ -axis, so the latus rectum is the vertical line $x = a$ . Substituting $x = a$ into the parabola's equation gives the $y$ -coordinates of the two endpoints:

y^2 = 4a(a) = 4a^2 \;\Rightarrow\; y = \pm 2a.

where: $a$ is the distance from the vertex to the focus, and $y = \pm 2a$ are the two points where the vertical line through the focus meets the curve. The endpoints are therefore $L(a, 2a)$ and $L'(a, -2a)$ , and the length of the chord joining them is simply the difference of their $y$ -values.

\boxed{\text{Length of latus rectum} = LL' = 2a - (-2a) = 4a}

Why the same $4a$ works for every orientation. The other three standard forms — $y^2 = -4ax$ (opens left), $x^2 = 4ay$ (opens up), $x^2 = -4ay$ (opens down) — are mirror images or rotations of $y^2 = 4ax$ , obtained by the same substitution trick: set the axis-variable equal to the focus coordinate and solve for the other. In every case the algebra produces the identical length, $4a$ , because $4a$ is always the coefficient sitting next to the linear variable in the standard equation. That gives a fast recall rule: whatever number multiplies the linear variable in the standard form is the latus rectum length — no re-derivation needed once the equation is in standard form.

Figure

Parabola y squared equals 4 a x with focus S and the vertical latus rectum chord LL prime through the focus

Latus rectum for the other three orientations.

$y^2 = -4ax$ (opens left): focus $(-a, 0)$ , endpoints $(-a, \pm 2a)$ , length $4a$ .
$x^2 = 4ay$ (opens up): focus $(0, a)$ , endpoints $(\pm 2a, a)$ , length $4a$ .
$x^2 = -4ay$ (opens down): focus $(0, -a)$ , endpoints $(\pm 2a, -a)$ , length $4a$ .

Why it matters. The latus rectum measures the "focal width" of the parabola. A larger $a$ pushes the focus farther from the vertex and makes the curve wider (flatter) near the focus; a smaller $a$ gives a narrower, more sharply curved parabola — a fact used, for instance, when comparing how tightly two reflector or antenna dishes curve near their focal point.

Derivation ALength of the latus rectum from the standard equation

The goal is to show, purely from the algebra of substituting the focus's $x$ -coordinate into the standard equation $y^2 = 4ax$ , that the chord through the focus perpendicular to the axis has length exactly $4a$ , matching the coefficient already visible in the equation, and to confirm the same reasoning transfers unchanged to the other three standard orientations.

Step 1. Locate the focus. For $y^2 = 4ax$ , comparing with the standard form tells us the focus is $S(a, 0)$ , a distance $a$ from the vertex along the axis of symmetry (the $x$ -axis here).

Step 2. Set up the chord. The latus rectum must pass through $S(a,0)$ and be perpendicular to the axis; since the axis is horizontal, this chord is the vertical line $x = a$ .

Step 3. Find where that line meets the parabola. Substitute $x = a$ into $y^2 = 4ax$ : $y^2 = 4a \cdot a = 4a^2$ , so $y = 2a$ or $y = -2a$ . These give the two endpoints $L(a, 2a)$ and $L'(a, -2a)$ .

Step 4. Compute the length. The chord is vertical, so its length is just the difference of the $y$ -coordinates: $LL' = 2a - (-2a) = 4a$ . This confirms the boxed result and shows the " $4a$ " is not a coincidence — it falls straight out of solving $y^2 = 4a \cdot a$ . Verified: the same four steps, run with $x=0$ or $y=0$ replaced by the correct focus coordinate, give length $4a$ for $y^2=-4ax$ , $x^2=4ay$ , and $x^2=-4ay$ as well.

Worked example

Problem. Find the length of the latus rectum and its endpoints for the parabola $y^2 = 20x$ . Then, separately, find the equation and focus of a downward-opening parabola (vertex at the origin) whose latus rectum has length $16$ .

Solution — Part A.

Step 1. Compare $y^2 = 20x$ with $y^2 = 4ax$ : matching coefficients, $4a = 20$ , so $a = 5$ .

Step 2. Length of latus rectum $= 4a = 20$ .

Step 3. The focus is $(a, 0) = (5, 0)$ , and the endpoints are $(a, 2a)$ and $(a, -2a)$ , i.e. $(5, 10)$ and $(5, -10)$ .

Step 4. Check: the distance between $(5,10)$ and $(5,-10)$ is $10 - (-10) = 20$ , matching $4a = 20$ . Answer: length $= 20$ ; endpoints $(5, 10)$ and $(5, -10)$ .

Solution — Part B.

Step 1. Downward opening with vertex at the origin means the standard form is $x^2 = -4ay$ .

Step 2. Latus rectum length $= 4a = 16 \Rightarrow a = 4$ .

Step 3. Equation: $x^2 = -4(4)y = -16y$ . Focus for this orientation is $(0, -a) = (0, -4)$ .

Step 4. Verify: endpoints of the latus rectum should be $(\pm 2a, -a) = (\pm 8, -4)$ . Substituting $(8, -4)$ into $x^2 = -16y$ : left side $= 64$ , right side $= -16(-4) = 64$ — consistent. Answer: equation $x^2 = -16y$ ; focus $(0, -4)$ .

Common traps

Confusing the length of the latus rectum ( $4a$ ) with the distance from vertex to focus ( $a$ ) — they differ by a factor of $4$ .
Forgetting the latus rectum must pass through the focus and be perpendicular to the axis — any other chord through the focus is not the latus rectum.
Sign errors for leftward/downward-opening parabolas — e.g. for $y^2 = -4ax$ , the focus is $(-a, 0)$ and the endpoints are $(-a, 2a)$ and $(-a, -2a)$ , not $(a, \ldots)$ .
Reading the coefficient itself as $a$ instead of solving $4a = \text{coefficient}$ first — for $y^2 = 20x$ , mistakenly taking $a = 20$ instead of $a = 5$ .
Reaching for the ellipse/hyperbola formula $\dfrac{2b^2}{a}$ — a parabola has no second axis length $b$ ; its latus rectum is simply $4a$ , not a ratio of two axis lengths.

Ask the tutor: Want to see why the latus rectum endpoints always come out as $(a, \pm 2a)$ — ask the AI tutor to derive it with you step by step, or to check your own worked example.

Why is the latus rectum always exactly 4a and never some other multiple of a?

It falls straight out of the algebra, not a rule to memorise blindly. Setting $x = a$ (the focus's own coordinate) into $y^2 = 4ax$ gives $y^2 = 4a^2$ , so $y = \pm 2a$ . The two endpoints are $2a$ apart on each side of the axis, and $2a - (-2a) = 4a$ . The " $4$ " is simply what appears when you double $2a$ to get the full chord length, and it will show up the same way for any of the four standard orientations.

Does the latus rectum formula change if the vertex is not at the origin?

No — the length $4a$ depends only on the coefficient $a$ in the standard equation, not on where the vertex sits. If the parabola is shifted, e.g. $(y-k)^2 = 4a(x-h)$ with vertex $(h,k)$ , the focus moves to $(h+a, k)$ and the latus rectum endpoints shift to $(h+a, k \pm 2a)$ , but the length itself is still $4a$ . Only the position of the chord changes with translation, never its length.

How is the latus rectum different from the directrix?

They are different objects entirely. The latus rectum is a chord — a line segment through the focus, perpendicular to the axis, with a finite length $4a$ . The directrix is a full line, $x = -a$ (for $y^2=4ax$ ), lying outside the curve on the opposite side of the vertex from the focus, used in the parabola's defining property that every point is equidistant from the focus and the directrix. One is a segment of length $4a$ sitting on the curve; the other is an infinite line that never touches the curve.

Parabola latus rectum

The core idea

Worked example

Common traps

Before this

Where this leads