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Hyperbola revision notes

Every idea in the Hyperbola chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 10 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 equation and its parts

Standard hyperbola
8× asked, last 2026
x2a2−y2b2=1:b2=a2(e2−1),c=ae=a2+b2\frac{x^2}{a^2}-\frac{y^2}{b^2}=1:\quad b^2=a^2(e^2-1),\qquad c=ae=\sqrt{a^2+b^2}
Transverse axis 2a2a (x-axis), conjugate 2b2b; e>1e>1. Vertices (±a,0)(\pm a,0), foci (±ae,0)(\pm ae,0), directrices x=±a/ex=\pm a/e.
xxyy(3,0)(3,0)(−3,0)(-3,0)y=baxy=\frac baxx=aex=\frac ae
a2=4a^2=4, b2=5b^2=5: c2=a2+b2=9c^2=a^2+b^2=9, foci (±3,0)(\pm3,0) outside the vertices; asymptotes are the diagonals of the 2a×2b2a\times2b box
  • Not c2=a2−b2c^2=a^2-b^2 (that is the ellipse). For a hyperbola c2=a2+b2c^2=a^2+b^2, and bb may be larger than aa. a2=4a^2=4, b2=5b^2=5 puts the foci at (±3,0)(\pm3,0), not (±1,0)(\pm1,0).
Latus rectum
19× asked, last 2026
LR=2b2a=2a(e2−1),ends (±ae, ±b2a)\text{LR}=\frac{2b^2}{a}=2a(e^2-1),\qquad \text{ends } \left(\pm ae,\ \pm\frac{b^2}{a}\right)
Focal chord perpendicular to transverse axis; length 2(semi-conjugate)2^2/(semi-transverse): 2a2/b2a^2/b for −x2/a2+y2/b2=1-x^2/a^2+y^2/b^2=1.
xxyyS(8,0)S(8,0)(8,143)(8,\frac{14}3)66
a=6a=6, b2=28b^2=28: the full chord through SS is 2b2a=283\frac{2b^2}a=\frac{28}3; 143\frac{14}3 is only half
  • Not b2/ab^2/a or 2b/a2b/a. The full length is 2b2/a2b^2/a; b2/ab^2/a is only the focus-to-end half. With a=6a=6, b2=28b^2=28: LR =28/3=28/3, not 14/314/3.
  • Quick method. Angle the latus rectum subtends: halve it by symmetry. At the centre tan⁡(θ/2)=(b2/a)/(ae)\tan(\theta/2)=(b^2/a)/(ae). At the other focus the base is 2ae2ae, so a right angle there means b2/a=2aeb^2/a=2ae.
Transverse axis along y
5× asked, last 2026
−x2a2+y2b2=1:e2=1+a2b2,foci (0,±be),LR=2a2b-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1:\quad e^2=1+\frac{a^2}{b^2},\qquad \text{foci }(0,\pm be),\qquad \text{LR}=\frac{2a^2}{b}
The positive term holds the transverse axis (2b2b here, on the y-axis). Directrices y=±b/ey=\pm b/e. RHS −1-1 means this form.
xxyy(0,5)(0,5)(0,−5)(0,-5)
x216−y29=−1\frac{x^2}{16}-\frac{y^2}9=-1 (red) opens up and down: e=53e=\frac53, foci (0,±5)(0,\pm5). Its conjugate (dashed) shares the asymptotes
  • x2/a2−y2/b2=−1x^2/a^2-y^2/b^2=-1 is the conjugate of the standard hyperbola: same asymptotes, axes swapped, and 1/e2+1/e′2=11/e^2+1/e'^2=1.
  • Not e2=1+b2/a2e^2=1+b^2/a^2 by habit. Divide by the square under the positive term. x2/16−y2/9=−1x^2/16-y^2/9=-1 opens up and down: e2=1+16/9e^2=1+16/9, so e=5/3e=5/3 and LR =32/3=32/3 (not e=5/4e=5/4).
Shifted hyperbola, centre (h, k)
4× asked, last 2026
(x−h)2a2−(y−k)2b2=1,foci (h±ae, k)\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1,\qquad \text{foci }(h\pm ae,\ k)
Centre = midpoint of the foci; 2ae2ae = distance between them. Shifting changes no length and not ee.
xxyy(1,14)(1,14)(1,−12)(1,-12)(1,6)(1,6)centre (1,1)(1,1)
Centre = midpoint (1,1)(1,1), c=13c=13; (1,6)(1,6) lies on the focal line, so it is a vertex: b=5b=5, a2=169−25=144a^2=169-25=144
  • A point of the hyperbola that lies on the line through the foci is a vertex. Foci (1,14)(1,14) and (1,−12)(1,-12), curve through (1,6)(1,6): centre (1,1)(1,1), c=13c=13; transverse axis on y, so b=5b=5, a2=169−25=144a^2=169-25=144.
  • Not swapped x2x^2, y2y^2 coefficients or the original size. The centroid of PP and both foci traces the hyperbola shrunk by 3 about its centre: same centre and orientation, a2a^2 and b2b^2 divided by 9.

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Practise Hyperbola

6 more concepts in 4 topics

Focus, directrix and focal distances; Eccentricity and data problems; Lines and the rectangular hyperbola; and Hyperbola with ellipses and lines. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Focus, directrix and focal distances

  • Focus–directrix definition5×
  • Focal distances6×

Eccentricity and data problems

  • Fixing a, b and e from given data12×

Lines and the rectangular hyperbola

  • Rectangular hyperbola3×

Hyperbola with ellipses and lines

  • Ellipse–hyperbola links12×
  • Line meeting a hyperbola4×

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