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Ellipse revision notes

Every idea in the Ellipse chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 16 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Standard ellipse

Major axis along x (a > b)
6× asked, last 2023
x2a2+y2b2=1 (a>b>0),b2=a2(1−e2),foci (±ae, 0)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ (a>b>0),\qquad b^2=a^2(1-e^2),\qquad \text{foci } (\pm ae,\,0)
0<e<10<e<1. Vertices (±a,0)(\pm a,0), minor-axis ends (0,±b)(0,\pm b); major axis 2a2a, minor axis 2b2b, foci 2ae2ae apart.
xxyyS(ae,0)S(ae,0)S′S'aabbx=aex=\frac ae
a=3a=3, b=2b=2: e=53e=\frac{\sqrt5}3, foci (±5,0)(\pm\sqrt5,0) inside, directrices x=±aex=\pm\frac ae outside
  • Directrices x=±a/ex=\pm a/e, 2a/e2a/e apart; a focus to its own directrix: a/e−ae=a(1−e2)/ea/e-ae=a(1-e^2)/e.
  • Not e=b/ae=b/a or e=1−b/ae=\sqrt{1-b/a}. The ratio inside the root is squared: e2=1−b2/a2e^2=1-b^2/a^2. With b/a=1/2b/a=1/2, e=3/2e=\sqrt3/2, not 1/21/2 or 1/21/\sqrt2.
Major axis along y (b > a)
9× asked, last 2026
x2a2+y2b2=1 (b>a),a2=b2(1−e2),foci (0,±be),ℓ=2a2b\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ (b>a),\qquad a^2=b^2(1-e^2),\qquad \text{foci } (0,\pm be),\qquad \ell=\frac{2a^2}{b}
Divide to get x2/A+y2/B=1x^2/A+y^2/B=1; the larger denominator is (semi-major)² and sets ee, foci, directrices y=±b/ey=\pm b/e.
xxyy(4,3)(4,3)(0,5)(0,5)(0,−5)(0,-5)
a2=20a^2=20, b2=45b^2=45: foci (0,±5)(0,\pm5); the latus rectum (red) is 2a2b=853\frac{2a^2}b=\frac{8\sqrt5}3
  • Major axis along y for x2/p+y2/q=1x^2/p+y^2/q=1 needs q>p>0q>p>0. If p=f(u)p=f(u), q=f(v)q=f(v) with ff strictly decreasing, the condition flips to v<uv<u.
  • Not e2=1−b2/a2e^2=1-b^2/a^2 when b>ab>a. That is negative. Use e2=1−a2/b2e^2=1-a^2/b^2 and ℓ=2a2/b\ell=2a^2/b. Major axis on y, through (4,3)(4,3), e=5/3e=\sqrt5/3: a2=20a^2=20, b2=45b^2=45, ℓ=85/3\ell=8\sqrt5/3.
Latus rectum
18× asked, last 2026
ℓ=2b2a=2a(1−e2),ends (±ae, ±b2a)\ell=\frac{2b^2}{a}=2a(1-e^2),\qquad \text{ends } \left(\pm ae,\ \pm\frac{b^2}{a}\right)
a>ba>b: the focal chord perpendicular to the major axis. The denominator is always the semi-major axis.
xxyyℓ=2b2a\ell=\frac{2b^2}a2b2b
The latus rectum (red) is the full focal chord ⊥\perp the major axis, 2b2a\frac{2b^2}a; not the minor axis 2b2b (dashed)
  • Not 2b2b (minor axis), 2a2/b2a^2/b (roles swapped) or b2/ab^2/a (only half). The full latus rectum is 2(semi-minor)2/(semi-major)2(\text{semi-minor})^2/(\text{semi-major}).
  • Quick method. Given ee and ℓ\ell: ℓ=2a(1−e2)\ell=2a(1-e^2) gives aa at once. e=2/3e=2/3, ℓ=10\ell=10: a=9a=9, b2=45b^2=45. If e=1/2e=1/\sqrt2, then b2=a2/2b^2=a^2/2 and ℓ=a\ell=a.
Finding e from a condition
8× asked, last 2026
b=k ae  ⇒  k2e2=1−e2,e2=11+k2b=k\,ae \;\Rightarrow\; k^2e^2=1-e^2,\qquad e^2=\frac{1}{1+k^2}
Write every length in aa and ee; aa cancels and leaves an equation in ee. Keep the root with 0<e<10<e<1.
  • Not mixing full and half lengths. 'minor axis =k×=k\times distance between foci' is 2b=k⋅2ae2b=k\cdot2ae. Likewise aeae is centre to one focus; the foci are 2ae2ae apart.

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12 more concepts in 4 topics

Directrix and shifted centre; Focal distances, parametric form and area; Lines, chords and circles; and Areas, loci and other conics. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Directrix and shifted centre

  • Focus–directrix definition4×
  • Shifted ellipse5×

Focal distances, parametric form and area

  • Focal distances8×
  • Parametric point and auxiliary circle4×
  • Area of an ellipse1×

Lines, chords and circles

  • Line meeting an ellipse8×
  • Chord with a given midpoint6×
  • Line touching an ellipse3×
  • Circle and ellipse meeting7×

Areas, loci and other conics

  • Triangles and quadrilaterals6×
  • Locus giving an ellipse5×
  • Ellipse with a hyperbola or parabola13×

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