Conjugate Hyperbola and Its Properties
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Direct answer
Swapping the sign between the two squared terms converts the hyperbola x^2/a^2 - y^2/b^2 = 1 into its conjugate x^2/a^2 - y^2/b^2 = -1: the transverse axis 2a of the first becomes the conjugate axis of the second, and vice versa. The pair shares its centre and the same two asymptotes y = ±(b/a)x, while the eccentricities obey 1/e1^2 + 1/e2^2 = 1, with e1^2 = 1 + b^2/a^2 and e2^2 = 1 + a^2/b^2. The foci of the original sit at (±√(a^2 + b^2), 0) and of the conjugate at (0, ±√(a^2 + b^2)), so all four foci lie on one circle x^2 + y^2 = a^2 + b^2. Taking the conjugate of the conjugate returns the original hyperbola.
What you must remember
- Definition pair: x^2/a^2 - y^2/b^2 = 1 and x^2/a^2 - y^2/b^2 = -1 are conjugates; the second's transverse axis lies along y with length 2b, its conjugate axis 2a.
- Common asymptotes: y = ±(b/a)x serve both curves — conjugacy is precisely "same asymptotes, transverse axes swapped".
- Eccentricity relation: 1/e1^2 + 1/e2^2 = 1; for the standard hyperbola e1 = √(1 + b^2/a^2), for its conjugate e2 = √(1 + a^2/b^2).
- Equal eccentricities force e = √2: the rectangular (equilateral) case a = b, where each curve is the reflection of the other in y = x.
- Foci and the focal circle: (±√(a^2 + b^2), 0) and (0, ±√(a^2 + b^2)) — four concyclic points on x^2 + y^2 = a^2 + b^2, each at distance √(a^2 + b^2) from the centre.
- Latus rectum swap: 2b^2/a for the original, 2a^2/b for the conjugate.
- Rectangular check: xy = c^2 has conjugate xy = -c^2 — same asymptotes (the coordinate axes), branches in opposite quadrants.
One worked eccentricity exchange
Given the hyperbola x^2/9 - y^2/16 = 1, read a = 3, b = 4, so c = √(9 + 16) = 5. Then e1 = 5/3 with foci (±5, 0), while the conjugate y^2/16 - x^2/9 = 1 has e2^2 = 1 + 9/16 = 25/16, so e2 = 5/4 with foci (0, ±5). Verify the reciprocal relation: 1/e1^2 + 1/e2^2 = 9/25 + 16/25 = 1. Notice the economy — c is the same number √(a^2 + b^2) for both curves, only its axis changes, which is why the four foci share one circle. Both curves run along the asymptotes y = ±4x/3, so a quick sketch shows two pairs of branches opening in perpendicular directions inside the same asymptotic cross. Whenever a question gives one eccentricity and asks for the other, skip re-derivation: 1/e2^2 = 1 - 1/e1^2 finishes it in a line, and if the answer equals the input, the hyperbola must be rectangular.
How the exam frames it
JEE Main asks for the eccentricity of the conjugate given the original (or its equation), the equation of the conjugate, and identification questions on shared asymptotes — all one-step work if the reciprocal relation is memorised. Advanced prefers the inverse direction: a hyperbola with prescribed asymptotes y = ±(b/a)x has equation x^2/a^2 - y^2/b^2 = λ for a nonzero parameter λ, and each λ pairs a curve with its conjugate (λ and -λ); passing-through-a-point questions resolve λ and then demand eccentricities of both members. The standard slips are assuming the conjugate shares the foci (it shares only the focal circle) and misremembering the eccentricity relation as e1^2 + e2^2 = 1, which produces impossible values; check that e2 > 1 always — every hyperbola, conjugate or not, has eccentricity above 1. Conjugate hyperbolas sit within the conic sections unit of the syllabus for both papers.
Frequently asked questions
What is the conjugate of x^2/a^2 - y^2/b^2 = 1?
x^2/a^2 - y^2/b^2 = -1, equivalently y^2/b^2 - x^2/a^2 = 1, with the transverse axis along the y-axis of length 2b.
How are the eccentricities of conjugate hyperbolas related?
1/e1^2 + 1/e2^2 = 1; if the two eccentricities are equal, both must equal √2, the rectangular case.
Do a hyperbola and its conjugate share their foci?
No — the foci swap axes: (±√(a^2 + b^2), 0) against (0, ±√(a^2 + b^2)), all four lying on the circle x^2 + y^2 = a^2 + b^2.
Which features do the two curves genuinely share?
The centre and the asymptote pair y = ±(b/a)x; nothing else — axes, foci, vertices and latera recta all swap roles.
When is a hyperbola identical in shape to its conjugate?
When a = b: both eccentricities equal √2, the rectangular hyperbola, and each curve is the reflection of the other in the line y = x.