DWG NO. 8.3 — Lesson 3 of 4

Hyperbolas

Unit 8: Conic Sections · ~25 min

Objective Write the standard-form equation of a hyperbola from its key features, and identify its vertices, foci, and asymptotes from a standard-form equation.

Lesson 8.1's ellipse fixed the sum of the distances from a point to two foci. A hyperbola fixes the difference of those two distances instead. That single change in operation — sum versus difference — is why an ellipse closes into one oval loop while a hyperbola splits into two separate, mirror-image branches that never meet.

Standard forms

Transverse axis	Standard form	Vertices	Foci	Asymptotes
Horizontal	(x−h)\(\frac{^{2}}{a^{2}}\) − (y−k)\(\frac{^{2}}{b^{2}}\) = 1	(h±a, k)	(h±c, k)	y = k ± (\(\frac{b}{a}\))(x−h)
Vertical	(y−k)\(\frac{^{2}}{a^{2}}\) − (x−h)\(\frac{^{2}}{b^{2}}\) = 1	(h, k±a)	(h, k±c)	y = k ± (\(\frac{a}{b}\))(x−h)

Whichever variable is positive tells you the transverse axis — the axis the two vertices (and the two branches) sit on. Unlike the ellipse, there's no rule that a > b here; a is always paired with the positive term.

The key contrast with ellipses Ellipse: c² = a² − b², so c < a — foci sit inside the vertices.
Hyperbola: c² = a² + b², so c > a — foci sit outside the vertices.

The asymptotes are two lines through the center that the branches approach but never touch, farther out from center. They act as guide rails for sketching the curve: draw a rectangle of width 2a and height 2b centered at (h, k), extend its diagonals, and the branches hug those diagonals as they run to infinity.

Hyperbola \(\frac{x^{2}}{4}\) − \(\frac{y^{2}}{2.25}\) = 1 — a=2, b=1.5, c=\(\sqrt{4+2.25}\)=2.5, asymptotes y=±0.75x

Worked Example 1 · Features from an equation

ProblemFind the vertices, foci, and asymptotes of (x−1)\(\frac{^{2}}{9}\) − (y+2)\(\frac{^{2}}{16}\) = 1.

1Center: (1, −2). The x-term is positive, so the transverse axis is horizontal: a²=9 → a=3, b²=16 → b=4.

2Vertices: (1±3, −2) = (4, −2) and (−2, −2).

3c² = a²+b² = 9+16 = 25 → c = 5. Foci: (1±5, −2) = (6, −2) and (−4, −2).

4Asymptotes: y = −2 ± (\(\frac{4}{3}\))(x−1).

Vertices (4,−2), (−2,−2) · Foci (6,−2), (−4,−2)

Worked Example 2 · Equation from vertices and foci

ProblemWrite the equation of the hyperbola with vertices (0, ±4) and foci (0, ±6).

1Vertices and foci lie on the y-axis, so the transverse axis is vertical and centered at the origin: a = 4, c = 6.

2b² = c² − a² = 36 − 16 = 20.

3\(\frac{y^{2}}{16}\) − \(\frac{x^{2}}{20}\) = 1.

\(\frac{y^{2}}{16}\) − \(\frac{x^{2}}{20}\) = 1

Guided practice

1.Find a, b, c, and the asymptotes of \(\frac{x^{2}}{4}\) − \(\frac{y^{2}}{9}\) = 1.
Show answer

a = 2, b = 3, c = \(\sqrt{4+9}\) = \(\sqrt{13}\). Asymptotes: y = ±(\(\frac{3}{2}\))x.
2.Find the vertices and foci of \(\frac{y^{2}}{25}\) − \(\frac{x^{2}}{11}\) = 1.
Show answer

a = 5 (vertical axis), c² = 25+11 = 36 → c = 6. Vertices (0, ±5), foci (0, ±6).
3.Write the equation of the hyperbola centered at the origin with vertices (±6, 0) and foci (±10, 0).
Show answer

a = 6, c = 10, b² = 100−36 = 64. \(\frac{x^{2}}{36}\) − \(\frac{y^{2}}{64}\) = 1.
4.Write the equation of the hyperbola with vertices (2, 1) and (2, −5), and foci (2, 3) and (2, −7).
Show answer

Center (2, −2) (midpoint of vertices), vertical axis, a = 3 (distance to vertex), c = 5 (distance to focus). b² = 25−9 = 16. (y+2)\(\frac{^{2}}{9}\) − (x−2)\(\frac{^{2}}{16}\) = 1.
5.Explain why a hyperbola's foci are always farther from the center than its vertices are.
Show answer

Because c² = a² + b², and b² > 0, c² is always strictly greater than a² — so c > a. (This is the opposite of the ellipse, where c² = a² − b² forces c < a.)

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