Pre-Calculus Conic sections

Hyperbolas (standard form, foci, asymptotes)

Q: How do you find the asymptotes of a hyperbola?

For x squared over a squared minus y squared over b squared equals 1, they are y equals plus or minus b over a x through the center.

Q: How do you find the foci of a hyperbola?

Use c squared equals a squared plus b squared; the foci lie a distance c from the center along the opening axis.

Q: How is a hyperbola different from an ellipse?

An ellipse sums distances to the foci; a hyperbola takes the difference and has two open branches.

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Hyperbolas

Texas Precalculus (TEKS) • Standard P.3(I) • Conic Sections

Hyperbolas is a topic in Conic Sections in the Texas Essential Knowledge and Skills (Precalculus, §111.42). It is aligned to Standard P.3(I), which requires students to write the equation of a hyperbola with center (h, k).

A hyperbola is the set of points whose distances to two foci differ by a constant, with two branches approaching straight asymptotes.

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Theory

A hyperbola is the set of points whose distances to two foci differ by a constant. It has two branches. Centered at the origin:

\[\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\ (\text{opens left/right}),\qquad \dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1\ (\text{opens up/down}).\]

The branches approach two straight asymptotes, and the foci sit a distance \(c\) from the center with

\[c^2=a^2+b^2.\]

The positive term opens the hyperbola. If \(x^2\) is positive it opens sideways; if \(y^2\) is positive it opens up and down.

A hyperbola's two branches approach its asymptotes.

Standard form, asymptotes, and foci.

Standard form, asymptotes, and focal distance:

\[\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1,\quad y=\pm\dfrac{b}{a}x,\quad c^2=a^2+b^2\]

x squared over a squared minus y squared over b squared equals 1; asymptotes y equals plus or minus b over a x; c squared equals a squared plus b squared

Hyperbola adds: \(c^2=a^2+b^2\), unlike the ellipse's subtraction.

How to analyze a hyperbola

Find the positive term to get the opening direction; \(a^2\) is under it.
Vertices: a distance \(a\) from the center along that axis.
Asymptotes: \(y=\pm\dfrac{b}{a}x\) (through the center).
Foci: \(c=\sqrt{a^2+b^2}\) from the center.

Example 1 — Vertices, foci, asymptotes

For \(\dfrac{x^2}{9}-\dfrac{y^2}{16}=1\), find the vertices, foci, and asymptotes.

Solution

\(a^2=9,\ b^2=16\), so \(a=3,\ b=4\); \(c^2=9+16=25\).

\(\text{vertices}\)	\(=\)	\((\pm 3,0)\)
\(\text{foci}\)	\(=\)	\((\pm 5,0)\)
\(\text{asymptotes}\)	\(:\)	\(y=\pm\dfrac{4}{3}x\)

vertices plus or minus 3, foci plus or minus 5, asymptotes plus or minus four thirds x

Example 2 — Opens vertically

Describe \(\dfrac{y^2}{4}-\dfrac{x^2}{9}=1\).

Solution

The \(y^2\) term is positive, so the hyperbola opens up and down.

\(\text{vertices}\)	\(=\)	\((0,\pm 2)\)
\(\text{asymptotes}\)	\(:\)	\(y=\pm\dfrac{2}{3}x\)

opens vertically, vertices at 0 comma plus or minus 2

Example 3 — Focal distance

Find the foci of \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\).

Solution

For a hyperbola \(c^2=a^2+b^2\) (add).

\(c^2\)	\(=\)	\(16+9=25\)
\(\text{foci}\)	\(=\)	\((\pm 5,0)\)

foci at plus or minus 5

Example 4 — Write the asymptotes

Give the asymptotes of \(\dfrac{x^2}{4}-\dfrac{y^2}{25}=1\).

Solution

Asymptotes are \(y=\pm\dfrac{b}{a}x\) with \(a=2,\ b=5\).

\(y\)

\(=\)

\(\pm\dfrac{5}{2}x\)

asymptotes are plus or minus five halves x

Common pitfalls

Hyperbola adds for \(c\): \(c^2=a^2+b^2\), not minus.

The positive term, not the larger one, opens the curve. Direction follows the sign, not the size.

Asymptote slope is \(\dfrac{b}{a}\) for a horizontal opening (and \(\dfrac{a}{b}\) forms appear if you mislabel) — keep \(a\) under the positive term.

Frequently asked questions

What is a hyperbola?

The set of points whose distances to two foci differ by a constant; it has two branches and a pair of asymptotes.

How do you find the asymptotes of a hyperbola?

For \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\), they are \(y=\pm\dfrac{b}{a}x\) through the center.

How do you find the foci of a hyperbola?

Use \(c^2=a^2+b^2\); the foci lie a distance \(c\) from the center along the opening axis.

How is a hyperbola different from an ellipse?

An ellipse sums distances to the foci (\(c^2=a^2-b^2\)); a hyperbola takes the difference (\(c^2=a^2+b^2\)) and has two open branches.

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Locus definitions of conic sections

Hyperbolas (standard form, foci, asymptotes)

Hyperbolas

📖 Theory

How to analyze a hyperbola

Common pitfalls

Frequently asked questions

Theory