The chord through the focus that measures how open the curve is

What to assume they know

• Balancing a point against a line, and the four equations that result — the definition of the parabola, the frame in which the focus is (a, 0) and the directrix x = −a, and the standard equation y² = 4ax
• That the parabola is symmetric about its axis, which §10.4.1's first observation states and this proof uses
• Reading a perpendicular distance from a point to a vertical line as a horizontal difference
• Substituting a value of one variable into an equation and solving for the other

What they should be able to do

• State Definition 3 and identify the three conditions a latus rectum must satisfy
• Locate the latus rectum on a drawing of any of the four standard parabolas
• Reproduce the chapter's proof that this chord measures 4a, naming the definition step and the symmetry step separately
• Derive the same length by substituting x = a into y² = 4ax, and say why the two routes must agree
• Read a, the focus, the directrix, the axis and the latus rectum off any of the four standard equations
• Explain why the latus rectum length is the absolute value of the linear coefficient in every standard form
• Apply the standard form to a physical parabola specified by a depth and a width

Where it usually goes wrong

• "The latus rectum is any chord through the focus." Definition 3 imposes three conditions at once: through the focus, perpendicular to the axis, and both endpoints on the curve. Drop the perpendicularity and the length is no longer determined — focal chords come in every length from 4a upwards.
• "You find it by substituting into the equation." You can, and the answer agrees, but that is not the chapter's argument and it is the weaker one. The printed proof never touches y² = 4ax. Teaching the substitution as the reason loses the fact that the length is fixed by the definition before any coordinate system exists.
• "AC = 2a because the picture looks like it." AC equals FM because ACMF is a rectangle, and FM equals 2a because that is how the frame was set up in §10.4.1. Both links have to be said out loud or the proof is a picture with letters on it.
• "a is the latus rectum." a is a quarter of it. The chain is: vertex to focus is a, directrix to focus is 2a, focus to either end of the latus rectum is 2a, and the whole latus rectum is 4a. Four lengths, all multiples of a, and students substitute one for another freely.
• "The latus rectum touches the directrix." It is parallel to the directrix and sits a distance 2a from it. Fig 10.18 draws both, which is why the figure is worth redrawing rather than describing.
• "A fractional a means the equation is wrong." Items 2, 5 and 6 of Exercise 10.2 all give fractional a and are perfectly ordinary. The coefficient is 4a, so a is a whole number only when the coefficient is a multiple of four.
• "Depth and diameter go straight into the formula." In Example 17 the depth is the x-coordinate and the half-diameter is the y-coordinate. Getting the two the wrong way round is the standard error, and the reason Fig 10.31 marks the depth along the axis.

Questions to check understanding

• Given a standard equation, state focus, axis, directrix and latus rectum length
• Given the latus rectum length and the orientation, write the equation
• Prove that the length of the latus rectum of y² = 4ax is 4a
• Compute the area enclosed by the vertex and the two endpoints of the latus rectum
• Reflector and mirror problems specified by depth and diameter, in both directions
• Short-answer: why is perpendicularity part of Definition 3

Examples worth working on the board

Values marked verified are worked out here from the chapter's printed data.

• Fig 10.17 (p. 185). A parabola opening right, its focus marked on the axis, and the latus rectum drawn as a vertical chord through the focus with the label written beside it inside the artwork.
• Fig 10.18 (p. 185) — the proof figure, and the four letters must be carried over exactly. A is the upper end of the latus rectum, B the lower end, F the focus at (a, 0), M the foot of the perpendicular from the focus to the directrix, and C the foot of the perpendicular from A to the directrix. The directrix runs vertically through M and C. Read off the page image; all five labels sit inside the drawing.
• The chapter's proof (p. 185), in three moves:
  • AF = AC, because A is on the parabola and AC is its distance to the directrix — this is Definition 2 and nothing else.
  • AC = FM = 2a, because ACMF is a rectangle and FM is the focus-to-directrix distance, fixed at 2a when the frame was set up in §10.4.1.
  • Therefore AF = 2a; the mirror symmetry about the x-axis gives FB = 2a as well; so AB = 4a. Verified as complete: no equation is solved at any point. The only inputs are the defining property, the frame's choice of 2a, and the symmetry the chapter had already stated.
• The algebraic cross-check. Verified by an added substitution: the latus rectum sits at x = a, so y² = 4a(a) = 4a² and y = ±2a. The half-chord is 2a and the whole chord 4a — the same answer. The comparison is worth attention: the synthetic route explains why the length is 2a on each side, and the algebraic route confirms it without explaining anything.
• The reading rule this establishes. Verified across all four standard forms: in y² = 4ax the coefficient of x is 4a, and 4a is the latus rectum; the same holds for the coefficient of y in x² = 4ay, and for the magnitudes of the negative forms. So the latus rectum can be read straight off the equation as the size of the linear coefficient, with no arithmetic at all.
• Example 5 (p. 185, Fig 10.19). The parabola y² = 8x. Verified: y is the squared variable, which puts the mirror line along the x-axis; x carries a positive coefficient, so the curve opens rightward; matching against y² = 4ax gives a = 2. Hence a focus at (2, 0), a directrix at x = −2, and a latus rectum of 8. Fig 10.19 draws the last two for this case.
• Exercise 10.2 items 1–6 (p. 186). Asked for each: focus, axis, directrix and the latus rectum length. Verified, all six by an added comparison with the standard forms:
  • y² = 12x: a = 3, focus (3, 0), directrix x = −3, axis the x-axis, latus rectum 12
  • x² = 6y: a = 3/2, focus (0, 3/2), directrix y = −3/2, axis the y-axis, latus rectum 6
  • y² = −8x: a = 2, opens left, focus (−2, 0), directrix x = 2, latus rectum 8
  • x² = −16y: a = 4, opens down, focus (0, −4), directrix y = 4, latus rectum 16
  • y² = 10x: a = 5/2, focus (5/2, 0), directrix x = −5/2, latus rectum 10
  • x² = −9y: a = 9/4, opens down, focus (0, −9/4), directrix y = 9/4, latus rectum 9 Note the pattern the set is built to expose: in every case the latus rectum is the size of the linear coefficient, while a is that number divided by four and is often a fraction.
• Miscellaneous Example 17 (p. 202, Fig 10.31). A parabolic mirror whose focus sits 5 cm out from its vertex, hollowed to a depth of 45 cm; asked for the span AB across its open end. Verified: a = 5 gives y² = 20x, and at x = 45, y² = 900 so y = ±30 and AB = 60 cm.
• Miscellaneous Exercise item 1 (p. 204). A parabolic reflector measuring 20 cm across its rim and 5 cm from rim plane to vertex; locate its focus. Verified: the rim point is (5, 10) measured from the vertex along the axis, so 100 = 4a(5) and 4a = 20, giving a = 5. The focus is 5 cm from the vertex — the same reflector as Example 17, read in the opposite direction, which is why the two make a good pair.
• Miscellaneous Exercise item 6 (p. 204). Asked how much area is enclosed by a triangle whose corners are the vertex of x² = 12y and the two ends of that curve's latus rectum. Verified: 4a = 12 so a = 3, the focus is (0, 3), the latus rectum runs from (−6, 3) to (6, 3), the base is 12 and the height 3, so the area is 18 square units. This is the item that makes the latus rectum a length you actually compute with.

Figures to have open

• Fig 10.18 (p. 185) redrawn with A, B, C, F and M named and the rectangle ACMF outlined. This is the proof; the outlined rectangle is the step the printed figure leaves the reader to see for themselves.
• Fig 10.17 (p. 185) redrawn with Definition 3's three conditions annotated onto the chord. Standard schematic.
• A comparison still showing the latus rectum beside two other focal chords of different lengths through the same focus — not a printed figure, and the only cheap way to show why perpendicularity is part of the definition.
• The four standard equations with linear coefficients highlighted. Standard schematic built from Fig 10.15.
• Fig 10.31 (p. 202) redrawn with the 45 cm depth marked along the axis and the half-distance marked perpendicular to it. The printed figure marks the depth across the top, which is easy to misread as a width.

Where this sits in the book

• NCERT Mathematics, Textbook for Class XI, Chapter 10 "Conic Sections", §10.4.2 Latus rectum (p. 185), including Definition 3 and the length argument
• Example 5 (p. 185); Exercise 10.2 items 1–6 (p. 186)
• Miscellaneous Example 17 (p. 202); Miscellaneous Exercise on Chapter 10 items 1 and 6 (p. 204)
• Figures: Fig 10.17, Fig 10.18 and Fig 10.19 (p. 185); Fig 10.31 (p. 202)
• Deliberate reference inside the chapter: the frame that fixes the focus-to-directrix distance at 2a is §10.4.1, p. 183, and the symmetry the proof uses is the first observation on p. 184
• The chapter's Summary (p. 205) restates Definition 3 and the length 4a
• Forward comparison inside the chapter: the ellipse gets its own latus rectum at §10.5.4 (p. 192), worked out by substituting rather than by appealing to the definition. The hyperbola's, at §10.6.3 (p. 200), is not worked out at all — one sentence hands the result over on the ellipse's authority, with no algebra and no figure. The latus rectum measured on an open curve, by the ellipse's own calculation is built around supplying what is missing there