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The latus rectum is a one-line formula, so it gets learned as one: 4a for a parabola, 2b²/a for an ellipse. The marks are lost elsewhere, when the ellipse is vertical, when the question gives the latus rectum and asks for the eccentricity, or when a hyperbola appears and the student wonders whether the formula changes. This post derives all three lengths in the same way and works examples for each.
Where is this in the syllabus?
NCERT's Class 11 chapter Conic Sections (2026-27 reprint) has a latus rectum subsection for each curve: 10.4.2 for the parabola, 10.5.4 for the ellipse and 10.6.3 for the hyperbola. Its definition for the parabola is "a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola". The JEE (Main) 2026 syllabus lists "equations of conic sections (parabola, ellipse and hyperbola) in standard forms" in Unit 10. The JEE (Advanced) 2026 brochure adds foci, directrices and eccentricity.
Where do the formulas come from?
One method works for all three: put the x-coordinate of the focus into the equation and solve for y.
Parabola y² = 4ax. The focus is (a, 0). At x = a, y² = 4a², so y = ±2a. Length 4a.
Ellipse x²/a² + y²/b² = 1, a > b. The focus is (c, 0) with c² = a² − b². At x = c, y² = b²(1 − c²/a²) = b² · b²/a², so y = ±b²/a. Length 2b²/a.
Hyperbola x²/a² − y²/b² = 1. The focus is (c, 0) with c² = a² + b². At x = c, y² = b²(c²/a² − 1) = b² · b²/a², so y = ±b²/a. Length 2b²/a again.
The ellipse and hyperbola share the formula because, in both, the step c²/a² ± 1 leaves b²/a².
Worked example 1: a downward parabola
x² = −20y. Compare with x² = −4ay: 4a = 20, so a = 5.
Focus (0, −5), directrix y = 5.
Latus rectum length 20, with ends at (±10, −5).
Check one end: x² = 100 and −20y = −20(−5) = 100.
Worked example 2: a horizontal and a vertical ellipse
4x² + 9y² = 144. Divide by 144: x²/36 + y²/16 = 1, so a = 6, b = 4.
c = √(36 − 16) = √20 = 2√5, foci (±2√5, 0).
Latus rectum = 2(16)/6 = 16/3, with ends (±2√5, ±8/3).
Check one end: 20/36 + (64/9)/16 = 5/9 + 4/9 = 1.
Now 25x² + 9y² = 225, which is x²/9 + y²/25 = 1. The larger denominator is under y², so the major axis is vertical with a = 5, b = 3. Latus rectum = 2(9)/5 = 18/5.
Worked example 3: a hyperbola
4x² − 5y² = 20. Divide by 20: x²/5 − y²/4 = 1, so a² = 5, b² = 4.
c² = 5 + 4 = 9, so c = 3 and the foci are (±3, 0).
Latus rectum = 2(4)/√5 = 8/√5 = 8√5/5 ≈ 3.58, with ends (3, ±4/√5).
Check: 9/5 − (16/5)/4 = 9/5 − 4/5 = 1.
How do reverse questions work?
Rewrite the latus rectum with eccentricity. For an ellipse, b² = a²(1 − e²), so 2b²/a = 2a(1 − e²).
Example: an ellipse whose latus rectum equals half its minor axis. Then 2b²/a = b, so a = 2b and e = √(1 − b²/a²) = √(1 − 1/4) = √3/2 ≈ 0.866.
Which slips cost marks?
Using 2a²/b for a vertical ellipse instead of keeping a as the semi-major axis.
Forgetting to divide through to get 1 on the right before reading a and b.
Using c² = a² − b² for a hyperbola. It is c² = a² + b².
Giving the semi-latus rectum (b²/a or 2a) when the full length is asked.
Writing a = 20 for x² = −20y. The 20 is 4a.
If you want to see the latus rectum change as an ellipse flattens, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine
Derive all three lengths from a blank page, as above.
Find the latus rectum and its end points for y² = 12x, x²/49 + y²/24 = 1 and x²/16 − y²/20 = 1.
Solve one reverse question: an ellipse with latus rectum 5 and major axis 10. Find e.
Check every end point by substituting it back into the equation.
For a full routine across the chapter, read our approach to conic sections; for focal chords and tangents, see the parabola post for JEE Advanced. Tonight, do step 1 and box the hyperbola line where c² = a² + b².
