Hyperbola
In short: A hyperbola is a two-branch curve made of all points whose distances to two fixed foci differ by a constant, with equations like x²/a² – y²/b² = 1 and asymptotes y = ±(b/a)x.
PrecalculusGeometryG-GPE.A.3
What is Hyperbola?
A hyperbola looks like two mirror-image bowls opening away from each other. Instead of adding the two distances to the foci, as an oval does, it fixes their difference, and that single change splits the curve into two separate branches.
For \(\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1\), the branches open up and down, with vertices \((0, \pm a)\) and foci \((0, \pm c)\), where \(c^2 = a^2 + b^2\). The branches approach the asymptotes \(y = \pm\frac{a}{b}x\). When the positive term is the \(x^2\) one, the branches open left and right.
How to spot it
- A hyperbola has an equation that subtracts one squared term from another and equals 1, for example x²/16 – y²/9 = 1.
- A graph shows two separate curves that approach crossing diagonal lines.
- A problem concerns signals timed from two stations, or a cooling tower silhouette.
- You are asked for asymptotes, vertices or foci of a curve with a minus sign between squares.
The method
- Rewrite the equation so that it equals 1; the positive squared term tells which way the branches open.
- Read \(a^2\) under the positive term and \(b^2\) under the negative one.
- Compute \(c = \sqrt{a^2 + b^2}\) for the foci, and put vertices at distance \(a\) from the center.
- Draw the central rectangle with half-sides \(a\) and \(b\), extend its diagonals as asymptotes, and sketch the hyperbola’s branches hugging them.
Hyperbola: solved example
Problem
Find the vertices, foci and asymptotes of \(9y^2 – 16x^2 = 144\).
Solution
Divide by 144: \(\frac{y^2}{16} – \frac{x^2}{9} = 1\). So \(a^2 = 16\), \(b^2 = 9\), and the branches open up and down. Then \(c = \sqrt{16 + 9} = 5\).
The vertices are \((0, \pm 4)\), the foci are \((0, \pm 5)\), and the asymptotes are \(y = \pm\frac{4}{3}x\).
Try it
Question 1
Find the foci of \(\frac{x^2}{25} – \frac{y^2}{144} = 1\).
Show the answer
\(c = \sqrt{25 + 144} = 13\), so the foci are \((\pm 13, 0)\).
Question 2
Give the asymptotes of \(\frac{x^2}{36} – \frac{y^2}{4} = 1\).
Show the answer
\(a = 6\) and \(b = 2\), so \(y = \pm\frac{b}{a}x = \pm\frac{1}{3}x\).
Hyperbola vs. Ellipse
The oval is a single closed loop, found with a plus sign between the squares and a constant sum of distances. This curve has two open branches, a minus sign and a constant difference. Memory aid: the plus-sign curve stays bounded, while the minus-sign curve runs off to infinity along diagonal lines.
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