The latus rectum measured on an open curve, by the ellipse's own calculation

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What to assume they know

Putting the centre at the origin, and reading the axes off the equation — the ellipse's standard equation and how to substitute a coordinate into it
The right triangle hidden in the figure, and the single number that sets the shape — eccentricity e = c/a and the relation a² = b² + c², from which the identity for e² is one rearrangement
Once c outgrows a, the ellipse's eccentricity and equation become the hyperbola's — the hyperbola's standard equation and c² = a² + b²
The chord through the focus that measures how open the curve is — the parabola's latus rectum 4a, proved from the definition alone, which is the contrast this topic is built against
Solving a quadratic in one unknown, needed for the reverse problems

What they should be able to do

State Definition 6 and Definition 9 and say what the two have in common
Locate a latus rectum on a drawing of an ellipse and of a hyperbola
Derive the ellipse's latus rectum length by substituting the focus's abscissa
Express e² in terms of a and b for the ellipse, and show the cancellation that removes e from the answer
Verify the hyperbola's formula independently, and identify the single sign that differs
Explain why the chapter is entitled to assert the hyperbola's result in one sentence
Contrast the algebraic route used here with the definition-based route used for the parabola
Compare the latus rectum against the major or transverse axis and say when it can be the longer of the two
Read a latus rectum off a given equation, and recover an equation from a given latus rectum

Where it usually goes wrong

"The hyperbola must have a different formula." It has the identical one, and the reason is the cancellation, not luck. Students who accept the chapter's one-sentence assertion without checking it cannot say why it is true, and cannot reconstruct it when they misremember which of e² and 1 is the larger.
"An ellipse has one latus rectum." It has two, one at each focus, and they are equal. Definition 6 permits either focus and Fig 10.26 draws both. The same holds for the hyperbola, with one chord on each branch.
"l = b²/a is the latus rectum." That is the half-chord. The symmetry step doubling it is easy to skip and gives an answer exactly half the truth.
"The latus rectum is always shorter than the main axis." True for every ellipse, false for hyperbolas with b > a. Exercise 10.4 item 2 is the counterexample and it is printed in the book.
"Because 4a works for the parabola, some multiple of a works here." The ellipse's and hyperbola's answers need both a and b. Only the parabola, which has a single shape parameter, can express it through one letter.
"c has to be computed first." It does not appear in the final formula at all. That is precisely what the cancellation buys, and it is worth pointing at explicitly because students reflexively find c before doing anything else.
"Negative roots can be kept if you take the modulus." In Example 16 the rejected root is discarded because a is a length in a geometric configuration, not because of a sign convention. The quadratic is honest; the geometry is what selects.

Questions to check understanding

Given an ellipse or hyperbola in standard form, state the latus rectum length — including equations needing division first
Given the foci and the latus rectum, find the equation, discarding the inadmissible root
Given the latus rectum and the eccentricity, recover a and b
Derive the length 2b²/a for the ellipse
Show that the same formula holds for the hyperbola, identifying the step where the sign differs
Short-answer: can a latus rectum be longer than the transverse axis, and for which curve

Examples worth working on the board

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

Fig 10.26 (p. 192). The ellipse on axes with F₁ and F₂ marked, and two latus recta drawn — one through each focus. A and B are the ends of the right-hand chord, C and D the ends of the left-hand one, and the label points at the chord itself from below. Read off the page image; all lettering is inside the artwork. The figure showing two chords rather than one is the visual answer to "which focus?" in Definition 6.
Definition 6 (§10.5.4, p. 192). A segment at right angles to the major axis, through either focus, with both ends on the ellipse. Note the phrase permitting either focus: there are two latus recta and they are equal in length.
The derivation (p. 192). Let the half-chord at the right focus have length l, so A has coordinates (c, l), which the chapter immediately rewrites as (ae, l) using c = ae. Verified by an added substitution: putting these into x²/a² + y²/b² = 1 gives e² + l²/b² = 1, so l² = b²(1 − e²).
The identity that finishes it (p. 192). Verified by an added rearrangement: e² = c²/a², and c² = a² − b², so e² = (a² − b²)/a² = 1 − b²/a². Therefore 1 − e² is exactly b²/a², and l² = b² × b²/a² = b⁴/a², giving l = b²/a. The ellipse's symmetry makes the other half equal, so the whole chord is 2b²/a.
Why the cancellation is the point. Verified: e entered the working through the endpoint's abscissa and left through the identity, so the final answer names only a and b — the two numbers already sitting in the equation's denominators. The chapter performs this and does not remark on it; it is what makes the result usable without ever computing c.
Definition 9 and the one-sentence hyperbola (§10.6.3, p. 200). The definition is the same shape with "transverse axis" in place of "major axis", and the chapter then states that the same length 2b²/a follows as it did for the ellipse, without doing the work.
Checking that sentence — this brief's own verification, since the chapter supplies none. For x²/a² − y²/b² = 1 the focus is again at x = ae, so e² − l²/b² = 1 and l² = b²(e² − 1). Here c² = a² + b², so e² = 1 + b²/a² and e² − 1 is b²/a². Verified: l² = b⁴/a² again, so l = b²/a and the full chord is 2b²/a. The single difference is that the ellipse subtracts b²/a² from 1 and the hyperbola adds it, and because the derivation needs only the size of the gap between e² and 1, the two land on the same answer. The chapter's one-sentence claim is therefore justified, and showing why is far better television than accepting it.
The contrast with the parabola (§10.4.2, p. 185). Verified as a genuine difference in method: the parabola's 4a is established without solving any equation — the endpoint's distance to the focus equals its distance to the directrix, and that distance is the focus-to-directrix separation. The ellipse and hyperbola have no directrix in this chapter, so no such argument is available and substitution is the only route. Three curves, two proof styles, and the reason is which definitional ingredients each curve has to hand.
Example 9 (pp. 192–193). x²/25 + y²/9 = 1. Verified: a = 5, b = 3, so the latus rectum is 2 × 9 / 5 = 18/5.
Example 14 (pp. 200–201). Two hyperbolas. (i) x²/9 − y²/16 = 1. Verified: a = 3, b = 4, c = √25 = 5, e = 5/3, foci (±5, 0), vertices (±3, 0), latus rectum 2 × 16 / 3 = 32/3. (ii) y² − 16x² = 16. Verified: divide by 16 to get y²/16 − x²/1 = 1, so a = 4, b = 1, c = √17, e = √17/4, foci (0, ±√17), vertices (0, ±4), latus rectum 2 × 1 / 4 = 1/2. Note the pair is chosen so that one has b > a and the other b < a.
Example 15 (p. 201). Foci (0, ±3), vertices (0, ±√11/2). Verified: a² = 11/4 and c = 3, so b² = 9 − 11/4 = 25/4; the equation clears to 100y² − 44x² = 275. The √ over the 11 is printed and the text layer drops it. Taken as 11/2 without the radical, a² would be 121/4 and b² would come out negative. Read this item from the page image only.
Example 16 (p. 201) — the reverse problem, and the best one in the section. Foci (0, ±12), latus rectum 36. Verified: c = 12 and 2b²/a = 36 gives b² = 18a; feeding that into the relation between c, a and b leaves the quadratic a² + 18a − 144 = 0, whose roots are 6 and −24. A length cannot be negative, so a = 6 and b² = 108, giving 3y² − x² = 108. The discarded root is the teaching moment: the algebra offers two answers and the geometry rejects one.
Exercise 10.3 items 1–9 (p. 195), latus rectum part only; the rest belongs to Putting the centre at the origin, and reading the axes off the equation. Verified: for x²/36 + y²/16 = 1 it is 16/3; for x²/4 + y²/25 = 1 it is 8/5; for x²/16 + y²/9 = 1 it is 9/2; for x²/25 + y²/100 = 1 it is 5; for x²/49 + y²/36 = 1 it is 72/7; for x²/100 + y²/400 = 1 it is 10; for 36x² + 4y² = 144 it is 4/3; for 16x² + y² = 16 it is 1/2; for 4x² + 9y² = 36 it is 8/3.
Exercise 10.4 items 1–6 (p. 202). Six hyperbolas, each listed below with the eccentricity and latus rectum I derived from it. All six verified:
- x²/16 − y²/9 = 1 — a = 4, b = 3, c = 5, e = 5/4, latus rectum 9/2
- y²/9 − x²/27 = 1 — a = 3, b² = 27, c = 6, e = 2, latus rectum 18
- 9y² − 4x² = 36 divides to y²/4 − x²/9 = 1 — a = 2, b = 3, c = √13, latus rectum 9
- 16x² − 9y² = 576 divides to x²/36 − y²/64 = 1 — a = 6, b = 8, c = 10, e = 5/3, latus rectum 64/3
- 5y² − 9x² = 36 divides to y²/(36/5) − x²/4 = 1 — a = 6/√5, b = 2, e = √14/3, latus rectum 4√5/3
- 49y² − 16x² = 784 divides to y²/16 − x²/49 = 1 — a = 4, b = 7, c = √65, latus rectum 49/2
Exercise 10.4 items 12 and 13 (p. 202) — the reverse problems, matching Example 16. Item 12: foci (±3√5, 0) and latus rectum 8. Verified: c² = 45 and b² = 4a, so 45 = a² + 4a and a = 5, giving x²/25 − y²/20 = 1. Item 13: foci (±4, 0) and latus rectum 12. Verified: c = 4 and b² = 6a, so 16 = a² + 6a and a = 2, giving x²/4 − y²/12 = 1. Both discard a negative root.
The comparison that only the hyperbola permits. Verified: for an ellipse b < a always, so 2b²/a is always less than 2a — the latus rectum is never longer than the major axis. For a hyperbola b may exceed a, and Exercise 10.4 item 2 is a printed instance: its transverse axis is 6 while its latus rectum is 18, three times as long. The chapter never sets these two facts against each other and it is the sharpest available demonstration that a > b is an ellipse theorem and not a general truth.

Figures to have open

Fig 10.26 (p. 192) redrawn with both latus recta drawn and all four endpoints named. A redraw showing only one chord contradicts Definition 6's wording.
A hyperbola with a latus rectum drawn on each branch. The chapter prints no figure at all for §10.6.3 — checked on the p. 200 page image — so this must be built, and its absence is exactly why students do not picture the hyperbola's version.
A two-column identity panel: the ellipse's e² = 1 − b²/a² beside the hyperbola's e² = 1 + b²/a², with the subsequent cancellation run in parallel to the same b⁴/a². Not printed, and the topic's central argument.
Exercise 10.4 item 2 drawn to scale, with the transverse axis and the latus rectum both marked, so the latus rectum is visibly three times the longer. Not printed.
The parabola's Fig 10.18 recalled alongside, for the method contrast in section 11.

Where this sits in the book

NCERT Mathematics, Textbook for Class XI, Chapter 10 "Conic Sections", §10.5.4 Latus rectum, including Definition 6 and the derivation (p. 192); §10.6.3 Latus rectum, including Definition 9 and the one-sentence assertion (p. 200)
Example 9 (pp. 192–193); Examples 14, 15 and 16 (pp. 200–201)
Exercise 10.3 items 1–9 (p. 195); Exercise 10.4 items 1–6, 12 and 13 (p. 202)
Figure: Fig 10.26 (p. 192). §10.6.3 prints no figure of its own
Deliberate reference inside the chapter: the parabola's latus rectum, proved by a different method, is §10.4.2, p. 185 — The chord through the focus that measures how open the curve is
Deliberate references inside the chapter: the relations the cancellation needs are §10.5.1, p. 188 for the ellipse and §10.6, p. 196 for the hyperbola
The chapter's Summary (pp. 205–206) restates Definitions 6 and 9 and the length 2b²/a for both curves, and prints no derivation of either