Putting the centre at the origin, and reading the axes off the equation

What to assume they know

• The right triangle hidden in the figure, and the single number that sets the shape — the relation a² = b² + c², which is the substitution that makes the derivation close, and the eccentricity e = c/a
• Two fixed points and a fixed total distance — Definition 4 and the meaning of 2a as the defining constant
• Isolating one square root, squaring, isolating the survivor, and squaring again
• Recognising a perfect square inside a radical, and knowing that pulling it out requires the base to be non-negative
• Solving a pair of linear equations in two reciprocal unknowns, as Example 13 and Exercise 10.3 items 19 and 20 need

What they should be able to do

• State the two frame decisions §10.5.3 makes and say what each one is worth
• Carry out the forward derivation from the defining sum to the standard equation, naming both squaring steps
• Identify the point at which a² − c² is replaced by b², and say which earlier result licenses it
• Reproduce the converse argument and state the two focal distances as functions of x
• Explain why the linearity of those distances makes the constant sum obvious
• Derive the bounding inequalities on x and y, and describe the rectangle they cut
• Read the orientation, a, b and the foci off any equation in standard form, including one not yet divided down to 1
• Fit a standard ellipse to two given points by solving for the reciprocals of the denominators
• Recognise a described locus as an ellipse and produce its standard equation

Where it usually goes wrong

• "a² always sits beneath x²." False, and Fig 10.24(b) is printed precisely to stop it. When the foci are on the y-axis, a² sits under y². Example 10 and Exercise 10.3 items 2, 4, 6, 7 and 8 are all this case.
• "The bigger denominator is a² — that's just the rule." It is a rule with a reason: a > b was proved in §10.5.1, so whichever denominator is bigger must be a². Students who learn the rule without the reason cannot recover it when the hyperbola breaks it two sections later, and it does break there.
• "9x² + 4y² = 36 has a = 9." The standard form has 1 on the right. Divide first, always. Three items in Exercise 10.3 exist to enforce this.
• "The converse is obvious, so skip it." It needs b² = a² − c² and it needs the radicand to be recognised as a perfect square. And the chapter takes √((a + cx/a)²) as a + cx/a without checking the base is non-negative — which it is, because |x| ≤ a and c < a, but the check is not printed. Skipping the converse means never learning that the focal distances are linear, which is the section's best idea.
• "The curve gets arbitrarily close to x = a without reaching it." It reaches it. The Discussion's bound is inclusive at both ends, and the curve touches all four sides of the rectangle. Read this off the page image, not the text layer.
• "Solving for a² and b² directly." Two conditions in a² and b² are nonlinear. Substituting u = 1/a² and v = 1/b² makes them linear, and that is the technique Example 13 and items 19 and 20 are teaching.
• "Any two points determine an ellipse." Two points give two equations for two unknowns only after the orientation and the centre are stipulated. Every item in this family stipulates both, and the resulting a² and b² still have to be checked against the stipulated orientation — as Example 13's a² > b² check shows.

Questions to check understanding

• Given a standard equation, state foci, vertices, both axis lengths and eccentricity — including equations needing division first
• Given vertices and foci, or an axis length and foci, write the equation
• Given two points and a stipulated major axis, find the equation
• Derive the standard equation from the definition
• Prove the converse: show any solution of the equation has focal distances summing to 2a
• Locus problems: a rod sliding between the axes, an elliptical arch
• Short-answer: why is the major axis the one with the larger denominator

Examples worth working on the board

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

• Fig 10.24 (p. 189), two panels sharing one caption. Read off the page image; every label is inside the artwork. Panel (a): the ellipse wide, vertices A and B on the x-axis, foci F₁(−c, 0) and F₂(c, 0), C and D on the y-axis, P(x, y) on the curve, equation x²/a² + y²/b² = 1 printed beneath. Panel (b): the ellipse tall, with (0, ±a) as vertices, (0, ±c) as foci and (±b, 0) on the x-axis, equation x²/b² + y²/a² = 1. Note carefully: in panel (b) it is a² that sits under y².
• Fig 10.25 (p. 189). Panel (a) redrawn as the working diagram, with the same labels plus the two focal segments to P(x, y) drawn in. This is the figure the derivation actually refers to.
• The frame (§10.5.3, pp. 188–189). O is the midpoint of F₁F₂ and becomes the origin; the ray from O through F₂ is the positive x-axis, the ray through F₁ the negative; the perpendicular at O is the y-axis. So F₁ is (−c, 0) and F₂ is (c, 0), and the definition reads PF₁ + PF₂ = 2a.
• The forward derivation (pp. 189–190). Both radicals are written out, one is moved to the right, and both sides are squared. Verified as necessary: if you square with both radicals on the left you get a cross term containing their product and are no better off. Isolating first is what makes the first squaring productive. The simplification leaves a single radical equal to a − (c/a)x, and a second squaring gives x²/a² + y²/(a² − c²) = 1.
• The substitution (p. 190). a² − c² is replaced by b², which is legitimate only because §10.5.1 proved a² = b² + c². Verified as the load-bearing step: without that earlier result the denominator has no name and the equation is not standard.
• The converse (p. 190), stated for 0 < c < a. From the equation, y² = b²(1 − x²/a²). Substituting into PF₁ and using b² = a² − c², verified by an added expansion: (x + c)² + (a² − c²)(a² − x²)/a² simplifies to a² + 2cx + c²x²/a², which is exactly (a + cx/a)². So PF₁ = a + (c/a)x, and symmetrically PF₂ = a − (c/a)x. Adding gives 2a, so the point satisfies the definition.
• The linearity, which the chapter does not comment on. Verified: both focal distances are straight-line functions of x with slopes +c/a and −c/a — that is, +e and −e. As a point slides right by one unit, one focal distance grows by e and the other shrinks by e. The sum cannot change. This is the single most video-worthy fact in the section and the chapter passes over it in one line.
• The Discussion (p. 191). Read off the page image, sign by sign: the chapter derives x²/a² = 1 − y²/b², notes it is at most 1, concludes x² ≤ a², and writes the bound as −a ≤ x ≤ a — inclusive at both ends. It then states the same for y between −b and b. So the curve lies inside the rectangle and touches all four of its sides.
• The observations (p. 191). The ellipse is symmetric in both axes, since changing the sign of x or of y or of both leaves the equation alone. And every focus sits on the major axis, which can be picked out by seeing which squared variable carries the larger denominator.
• Example 9 (pp. 192–193). x²/25 + y²/9 = 1. Verified: 25 > 9, so the major axis is horizontal; a = 5, b = 3, c = 4; foci (±4, 0); vertices (±5, 0); major axis 10, minor axis 6, e = 4/5.
• Example 10 (p. 193). 9x² + 4y² = 36. Verified: divide by 36 to get x²/4 + y²/9 = 1; now 9 > 4 and the larger denominator sits under y², so the major axis is vertical, a = 3, b = 2, c = √5, foci (0, ±√5), vertices (0, ±3), e = √5/3. The division is the whole point of the item.
• Example 11 (pp. 193–194). Vertices (±13, 0), foci (±5, 0). Verified: a = 13, c = 5, b = 12; equation x²/169 + y²/144 = 1.
• Example 12 (p. 194). Major axis 20, foci (0, ±5). Verified: the foci on the y-axis force the vertical orientation, so the form is x²/b² + y²/a² = 1 with a = 10 and c = 5, giving b² = 75 and the equation x²/75 + y²/100 = 1.
• Example 13 (pp. 194–195). Major axis along the x-axis, through (4, 3) and (−1, 4). Verified by an added elimination: writing u = 1/a² and v = 1/b² makes the two conditions linear — 16u + 9v = 1 and u + 16v = 1 — giving v = 15/247 and u = 7/247, so a² = 247/7 and b² = 247/15, and the equation clears to 7x² + 15y² = 247. Verified as consistent: a² ≈ 35.3 exceeds b² ≈ 16.5, so the major axis really is horizontal as the question stipulated. The reciprocal substitution is the technique; without it the system is nonlinear.
• Exercise 10.3 items 1–9 (p. 195), asking for foci, vertices, both axis lengths, eccentricity and latus rectum:
  • x²/36 + y²/16 = 1 — verified: a = 6, b = 4, c = 2√5, e = √5/3
  • x²/4 + y²/25 = 1 — verified: vertical, a = 5, b = 2, c = √21
  • x²/16 + y²/9 = 1 — verified: a = 4, b = 3, c = √7
  • x²/25 + y²/100 = 1 — verified: vertical, a = 10, b = 5, c = 5√3
  • x²/49 + y²/36 = 1 — verified: a = 7, b = 6, c = √13
  • x²/100 + y²/400 = 1 — verified: vertical, a = 20, b = 10, c = 10√3
  • 36x² + 4y² = 144 — verified: divides to x²/4 + y²/36 = 1, vertical, a = 6, b = 2
  • 16x² + y² = 16 — verified: divides to x²/1 + y²/16 = 1, vertical, a = 4, b = 1
  • 4x² + 9y² = 36 — verified: divides to x²/9 + y²/4 = 1, horizontal, a = 3, b = 2 The latus rectum part of these items belongs to The latus rectum measured on an open curve, by the ellipse's own calculation.
• Exercise 10.3 items 10–20 (p. 195), asking for the equation: vertices (±5, 0) with foci (±4, 0); vertices (0, ±13) with foci (0, ±5); vertices (±6, 0) with foci (±4, 0); major-axis ends (±3, 0) and minor-axis ends (0, ±2); major-axis ends (0, ±√5) and minor-axis ends (±1, 0); major axis 26 with foci (±5, 0); minor axis 16 with foci (0, ±6); foci (±3, 0) with a = 4; b = 3 and c = 4 with a centred, x-axis-focused ellipse; a centred ellipse whose major axis runs vertically, through (3, 2) and (1, 6); and one whose major axis runs horizontally, through (4, 3) and (6, 2). Verified, the two hardest: item 19 gives 9p + 4q = 1 and p + 36q = 1 with p = 1/b² and q = 1/a², so q = 1/40 and p = 1/10, giving x²/10 + y²/40 = 1 — and a² = 40 does exceed b² = 10, consistent with the stipulated vertical major axis. Item 20 gives 16u + 9v = 1 and 36u + 4v = 1, so u = 1/52 and v = 1/13, giving x²/52 + y²/13 = 1. Item 14's √5 is printed under a radical sign; the text layer drops it. Take it from the page image.
• Miscellaneous Example 19 (pp. 203–204, Fig 10.33). A straight rod 15 cm long leans with its end A touching the x-axis and its end B the y-axis; a point P is marked on it 6 cm from A. Asked: show P traces an ellipse. Verified: PB is 9 cm; dropping perpendiculars gives cos θ = x/9 from the upper triangle and sin θ = y/6 from the lower, so squaring and adding gives x²/81 + y²/36 = 1. The identity cos²θ + sin²θ = 1 is what eliminates the angle, and it is a Chapter 3 result used here without re-derivation.
• Miscellaneous Exercise item 5 (p. 204). A rod of length 12 cm with its ends on the axes, and P at 3 cm from the end touching the x-axis. Verified by the same method: the segment to the y-axis end is 9 cm, so x²/81 + y²/9 = 1.
• Miscellaneous Exercise item 4 (p. 204). A semi-elliptical arch spanning 8 m and standing 2 m tall above its midpoint; asked how high it stands at a place 1.5 m in from one end. Verified: a = 4 and b = 2, and 1.5 m from an end is 2.5 m from the centre, so y² = 4(1 − 6.25/16) = 2.4375 and the height is about 1.56 m.

Figures to have open

• Fig 10.24 (p. 189) both panels, redrawn side by side with the denominators colour-matched to the axes. The single most useful still in the topic, because the a-under-y² case is where most errors live.
• Fig 10.25 (p. 189) redrawn as the working diagram with both focal segments to a general point.
• The sliding-point movement for section 9 — two bars, opposite slopes, constant total. Not a printed figure and the topic's whole argument.
• The bounding rectangle round the ellipse with four touch points. The chapter describes four lines in prose (p. 191) and draws no such figure.
• A reciprocal-substitution panel for Example 13: the nonlinear pair beside the linear pair it becomes. Standard schematic.
• Fig 10.33 (pp. 203–204) redrawn with the rod, both perpendiculars, and the two angles marked. The printed figure marks θ in two places and it is easy to lose.

Where this sits in the book

• NCERT Mathematics, Textbook for Class XI, Chapter 10 "Conic Sections", §10.5.3 Standard equations of an ellipse (pp. 188–191), including the derivation, the converse, the Discussion, the Note bounding the section, and the two observations
• Examples 9 to 13 (pp. 192–195); Exercise 10.3 (p. 195)
• Miscellaneous Example 19 (pp. 203–204); Miscellaneous Exercise on Chapter 10 items 4 and 5 (p. 204)
• Figures: Fig 10.24 and Fig 10.25 (p. 189); Fig 10.33 (p. 203)
• Deliberate backward reference inside the chapter: a² = b² + c² is §10.5.1, p. 188 — The right triangle hidden in the figure, and the single number that sets the shape
• Deliberate reference outside the chapter: the Pythagorean identity used in Miscellaneous Example 19 is a Chapter 3 result, used here without re-derivation
• The chapter's Summary (p. 205) prints only the horizontal standard form; the vertical one, the converse and the Discussion do not appear there