Hyperbola and Ellipse: JEE Main Complete Guide

Ellipse and Hyperbola together contribute 2–3 questions per JEE Main session from the Conic Sections chapter. These two conics share many structural parallels — both have two foci, both have standard tangent and normal formulas with the same form, and both have a director circle. However, the sign differences between their equations (plus vs. minus) lead to fundamentally different geometric properties, and it is these differences that JEE Main most frequently tests. This guide covers both conics systematically with the exact results needed for JEE Main.

Ellipse: Standard Forms and Key Properties

Standard ellipse: x²/a² + y²/b² = 1, with a greater than b (major axis along x-axis). b² = a²(1−e²), so e = sqrt(1 − b²/a²). Eccentricity 0 less than e less than 1 for ellipse. Foci: (±ae, 0). Directrices: x = ±a/e. Vertices: (±a, 0) and (0, ±b). Latus rectum: 2b²/a. For a vertical ellipse (b greater than a): swap roles of a and b. Focal distances: |PF1| + |PF2| = 2a (sum of distances from any point to both foci = 2a — the defining property of an ellipse). Auxiliary circle: x² + y² = a² — any point on the ellipse corresponds to a point on the auxiliary circle at the same x-value. Parametric form: x = a·cos(theta), y = b·sin(theta) (not the same theta as the eccentric angle in some texts — be consistent with your textbook convention). For the parabola comparisons and conic sections overview, see our Conic Sections Complete Guide.

Tangent to x²/a² + y²/b² = 1 at point (x1,y1): x·x1/a² + y·y1/b² = 1. In slope form: y = mx ± sqrt(a²m² + b²). Condition for y = mx + c to be tangent: c² = a²m² + b². Point of tangency with slope m: (−a²m/c, b²/c). Normal at (x1,y1): a²x/x1 − b²y/y1 = a² − b² = a²e². Director circle of ellipse: x² + y² = a² + b² (tangents from any point on this circle are perpendicular). Chord of contact from external point (h,k): x·h/a² + y·k/b² = 1. These formulas mirror the parabola formulas with appropriate substitution — the T=0 and T=S1 approach works for all conics.

Hyperbola: Standard Forms and Asymptotes

Standard hyperbola: x²/a² − y²/b² = 1. Eccentricity e = sqrt(1 + b²/a²) greater than 1. Foci: (±ae, 0). Vertices: (±a, 0). The hyperbola has two branches (right and left). Asymptotes: y = ±(b/a)x — lines that the hyperbola approaches but never reaches. For rectangular hyperbola: a = b, so asymptotes are y = ±x (at 90° to each other). Rectangular hyperbola in rotated form: xy = c² (a standard form tested in JEE Main). Parametric form for rectangular hyperbola: x = ct, y = c/t. Focal distances: |PF1| − |PF2| = ±2a (difference of distances from any point to the two foci = 2a). Key difference from ellipse: the difference of focal distances is constant for hyperbola, while the sum is constant for ellipse. Practise ellipse and hyperbola problems on our JEE Main math mock tests with a difficulty mix matching recent JEE Main sessions.

Tangent to x²/a² − y²/b² = 1 at point (x1,y1): x·x1/a² − y·y1/b² = 1. In slope form: y = mx ± sqrt(a²m² − b²). Condition for tangency: c² = a²m² − b² (note the minus sign vs. ellipse). The asymptotes themselves satisfy c² = a²m² − b² with c = 0 and m = ±b/a — confirming they are "tangents at infinity." Conjugate hyperbola: −x²/a² + y²/b² = 1 — shares the same asymptotes as the original hyperbola but has its transverse axis along the y-axis. Auxiliary circle for hyperbola: x² + y² = a².

Focal Chords, Normals, and Special Points

For ellipse x²/a² + y²/b² = 1, normal at (a·cos(theta), b·sin(theta)): (a·x)/cos(theta) − (b·y)/sin(theta) = a² − b². The normal passes through the focus if the point is an end of the minor or major axis — for a general point, four normals can be drawn from an external point to an ellipse, and their feet satisfy a certain quartic. JEE Main typically does not ask about all four normals but does ask about the normal at a given parametric point. Pole and polar for ellipse: polar of (h,k) with respect to x²/a² + y²/b² = 1 is x·h/a² + y·k/b² = 1 (same as chord of contact). For hyperbola: polar of (h,k) with respect to x²/a² − y²/b² = 1 is x·h/a² − y·k/b² = 1.

Rectangular hyperbola xy = c²: tangent at (ct, c/t) is x/t + yt = 2c (or x + yt² = 2ct). Normal: xt³ − yt − ct⁴ + c = 0. Chord joining (ct1, c/t1) and (ct2, c/t2): x + y·t1·t2 = c(t1+t2). For a focal chord of xy = c²: t1·t2 = −1 (same as parabola!). The asymptotes of xy = c² are the coordinate axes (x = 0 and y = 0). JEE Main tests rectangular hyperbola extensively because its simpler form allows rapid calculations that test the same conceptual understanding as the general hyperbola.

Common Mistakes and Exam Strategy

The most frequent JEE Main errors in ellipse and hyperbola: (1) Confusing the c² condition for tangency (c² = a²m² + b² for ellipse; c² = a²m² − b² for hyperbola — the sign matters). (2) Forgetting that for a hyperbola, b² = a²(e²−1) (not a²(1−e²) as in ellipse). (3) Using the wrong focal chord property for hyperbola vs. ellipse (difference vs. sum of focal distances). (4) Asymptotes of xy = c² are coordinate axes — some students forget this and try to find asymptotes using the formula for x²/a² − y²/b² = 1. Build a comparison table: for each property (eccentricity, foci, latus rectum, director circle, tangent condition, focal property), write the ellipse result and hyperbola result side by side. This comparison table is the most effective revision tool for this chapter pair. Sign up on our platform for 200+ ellipse and hyperbola practice problems. Our premium subscription includes conic sections topic tests with difficulty calibration. For the parabola that completes the conic sections family, see our Parabola Complete Guide.

Time management: JEE Main ellipse/hyperbola problems that directly ask you to apply the tangent or chord of contact formula take 90 seconds. Problems that require finding the intersection of a tangent with the directrix or establishing a geometric property take 3–4 minutes. Practise both types and allocate time accordingly — do not spend 5 minutes on a tangent formula question that should take 90 seconds.