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Precalculus

Hyperbola Properties

Enter the center (h, k), the semi-transverse axis a and the semi-conjugate axis b for the standard horizontal-transverse hyperbola (x − h)²/a² − (y − k)²/b² = 1. The calculator returns the vertices, the foci with c = √(a² + b²), the asymptote equations and the eccentricity.

Hyperbola Properties

Vertices, foci, asymptotes and eccentricity of a hyperbola.

Try:
Answercenter (0, 0), vertices (-3, 0) & (3, 0), foci (-5, 0) & (5, 0), e = 1.66667
  1. Equation(x − 0)²/9 − (y − 0)²/16 = 1
  2. Center(0, 0)
  3. Orientationhorizontal transverse axis (x-term positive)
  4. Vertices(-3, 0), (3, 0)
  5. Focic = √(9 + 16) = 5 → (-5, 0), (5, 0)
  6. Asymptotesy − 0 = ±(1.33333)·(x − 0)
  7. Eccentricitye = c/a = 1.66667
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Frequently asked questions

What are the asymptotes of a hyperbola?

Two straight lines the hyperbola approaches but never touches as |x| grows: y − k = ±(b/a)(x − h) for the horizontal-transverse case.

How does the formula for c differ between ellipse and hyperbola?

Ellipse: c² = a² − b². Hyperbola: c² = a² + b². In an ellipse the foci sit inside the curve; in a hyperbola they sit outside the vertices.

Is the eccentricity always greater than 1?

Yes. For a hyperbola e = c/a > 1, which captures the fact that the foci lie beyond the vertices.

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