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NCERT defines an ellipse in one sentence: the set of points whose distances from two fixed points add to a constant. Most students memorise the equation and forget the sentence, yet many questions about foci are faster with the sentence. This post shows why the constant is 2a, why the two distances are a − ex and a + ex, and works six focal-distance questions on one ellipse.
Is this in NCERT and the JEE syllabus?
The NCERT Class 11 chapter Conic Sections (reprint 2026-27) gives the definition, derives c² = a² − b² from the end of the minor axis, and derives the standard equation. In the converse step it shows that the distances to the foci are a + (c/a)x and a − (c/a)x. The NTA Syllabus for JEE (Main) 2026 lists the ellipse "in standard forms". Check the documents for your exam year.
Why is the constant 2a?
Take P at the end of the major axis, (a, 0). Its distances to the foci (c, 0) and (−c, 0) are a − c and a + c, which add to 2a. The sum is the same for every point, so it is 2a everywhere: the length of the major axis.
Now take the end of the minor axis, (0, b). Both distances are equal, so each is a. That right triangle gives b² + c² = a², which is where c² = a² − b² comes from. A useful picture: from either focus to the end of the minor axis is exactly a.
NCERT adds a condition: the constant must be greater than the distance between the foci. If it equals 2c, the points fill only the segment between the foci; if it is less, there are none.
Why are the distances a − ex and a + ex?
Put y² = b²(1 − x²/a²) into the distance formula and use b² = a² − c². NCERT shows the square root simplifies to a + (c/a)x. With e = c/a:
PS = a − ex (to the focus S at (ae, 0)),
PS′ = a + ex (to S′ at (−ae, 0)).
Both are linear in x, and they add to 2a. For a vertical ellipse, replace x by y.
Worked example: x²/100 + y²/64 = 1
Here a = 10, b = 8, c = √(100 − 64) = 6 and e = 0.6. The foci are S(6, 0) and S′(−6, 0).
Distances at x = 5. y² = 64(1 − 25/100) = 48, so y = 4√3. PS = 10 − 0.6 × 5 = 7 and PS′ = 13. Directly: √(1 + 48) = 7 and √(121 + 48) = 13. They add to 20 = 2a.
Perimeter of triangle PSS′. For any P on the ellipse off the major axis, it is 2a + 2c = 20 + 12 = 32.
Range of PS. From a − c = 4 (nearer vertex) to a + c = 16 (farther vertex).
Product PS · PS′. (a − ex)(a + ex) = a² − e²x². It is largest at x = 0, where both distances are 10 and the product is 100 = a². It is smallest at the vertices, 4 × 16 = 64 = b².
Inside or outside? For Q(6, 6): QS = 6 and QS′ = √(144 + 36) ≈ 13.42, a sum of about 19.42, less than 20, so Q is inside. Check: 36/100 + 36/64 = 0.9225 < 1.
Draw it. Fix two pins 12 cm apart and tie a 20 cm string to them. A pencil holding the string taut traces this ellipse in centimetres, because the two distances always add to the string's length.
How do you use 2a in a locus question?
Find the locus of a point whose distances from (0, 4) and (0, −4) add to 10. The foci are on the y-axis with c = 4, and 2a = 10, so a = 5 and b² = 25 − 16 = 9. The locus is x²/9 + y²/25 = 1. Check at (3, 0): each distance is √(9 + 16) = 5, which add to 10.
Does a real orbit follow the rule?
A planet's orbit is an ellipse with the Sun at one focus, so its nearest and farthest distances are a − c and a + c. NASA's Earth fact sheet gives perihelion 147.095 and aphelion 152.100 million km. They add to 299.195, so a ≈ 149.60 million km, matching the listed semimajor axis of 149.598. Half the difference gives c ≈ 2.50, and e = c/a ≈ 0.0167, the listed eccentricity.
Which mistakes cost marks?
Swapping signs. a − ex goes with the focus on the same side as P. Check at a vertex.
Taking 2a as the distance between foci. That is 2c.
Forgetting a vertical ellipse uses a ± ey.
Mixing up conics. The hyperbola has c² = a² + b², and the difference of its focal distances is 2a.
The same ellipse with tangents and the auxiliary circle is in our ellipse guide for JEE, and the curve's history is in our post on Apollonius of Perga.
A practice routine
Derive PS = a − ex from the distance formula once, without notes.
For x²/25 + y²/16 = 1, find both focal distances at x = 0, 3 and 5. (Answers: 5 and 5; 3.2 and 6.8; 2 and 8.)
Find the locus of points whose distances from (±2, 0) add to 6. (Answer: x²/9 + y²/5 = 1.)
Mix five ellipse questions with five on the parabola and hyperbola, so you must choose the right property each time.
When a definition like eccentricity still feels abstract, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
Today, check question 1 of the worked example with the distance formula yourself, then do the x²/25 + y²/16 = 1 version.
