The Hyperbola

Try this

Drag $P$ along the right branch. Does $∣ P F_{1} ∣ - ∣ P F_{2} ∣$ change? Drag $P$ onto the left branch: what is the difference now?
Drag $P$ far out and watch the “distance to the asymptote”. Does it ever become $0$ ?
The dashed box has length $2 a$ and width $2 b$ . How are the two asymptotes related to the diagonals of the box?
Read off $c$ , then measure half the diagonal of the box, $a^{2} + b^{2}$ . Increase $b$ : how do the opening and the eccentricity $e$ change?

The locus of points whose distances to two fixed points $F_{1}, F_{2}$ have a difference whose absolute value is a constant $2 a$ (with $0 < 2 a < ∣ F_{1} F_{2} ∣$ ) is called a hyperbola:

∣ P F_{1} ∣ - ∣ P F_{2} ∣ = 2 a

Without the absolute value, “ $= 2 a$ ” is the branch closer to $F_{2}$ , and “ $= - 2 a$ ” is the branch closer to $F_{1}$ .

Predict first

A point $P$ on the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ is at distance $10$ from the left focus $F_{1}$ . How far is $P$ from the right focus $F_{2}$ ?

only

4

only

16

4

16

cannot be determined

Choose an answer first; then you can see the result.

Why

The derivation is almost word for word the same as for the ellipse: square twice. The only difference is in the last step: here $c > a$ .

From the definition to the standard equation, then to the asymptotes

Step 1 / 6

Write down the definition
The foci are $F_{1} (- c, 0)$ and $F_{2} (c, 0)$ ; in the figure $a = 3$ . A point $P$ on the right branch satisfies $∣ P F_{1} ∣ - ∣ P F_{2} ∣ = 2 a$ .

$(x + c)^{2} + y^{2} - (x - c)^{2} + y^{2} = \pm 2 a$
Square twice
This time c > a
Draw a box
The asymptotes
Eccentricity

Two small tricks:

A hyperbola whose transverse axis is the $y$ -axis is $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ . Whichever term is positive tells you which axis is the transverse axis; this is different from the ellipse, where you compare the denominators.
Asymptotes: replace the $1$ on the right by $0$ , and $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 0$ factors into two lines. Conversely, every hyperbola with asymptotes $y = \pm \frac{b}{a} x$ can be written as $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = λ$ ( $λ \neq = 0$ ). The one with $λ = - 1$ is called the conjugate hyperbola; it shares the same box with the original.

Practice

Exercise 1Basic

Find the eccentricity of the hyperbola $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ (as a fraction).

Exercise 2Basic

One asymptote of the hyperbola $\frac{x ^{2}}{4} - \frac{y ^{2}}{12} = 1$ makes an acute angle with the positive $x$ -axis. How many degrees is it?

Exercise 3Advanced

A hyperbola has its foci on the $x$ -axis, focal distance $10$ and asymptotes $y = \pm \frac{1}{2} x$ . Find its $a^{2}$ .

Exercise 4Challenge

$F_{1}, F_{2}$ are the foci of the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ , and $P$ is on the hyperbola with $∠ F_{1} P F_{2} = 6 0^{\circ}$ . Find $∣ P F_{1} ∣ \cdot ∣ P F_{2} ∣$ .

Try this

Why

From the definition to the standard equation, then to the asymptotes

Practice

Mistake Museum · Exhibits from this stop

Getting the asymptote slopes upside down

Finding a hyperbola's foci with the ellipse relation