Hyperbolas Advanced. This lesson goes beyond core Algebra II. You can skip it.

The distance condition

Fix two points $F_1$ and $F_2$, the foci, and fix a positive constant. A hyperbola is the set of points $P$ whose distances to the two foci have that constant absolute difference:

Two details in that sentence do all the work.

The constant is written $2a$ rather than $k$ purely for convenience, exactly as with the ellipse. The algebra below turns it into a clean $a^2$, and $a$ will turn out to be the distance from the center to each turning point of the curve.

The absolute value is what creates two branches. Drop it and you are left with $PF_1 - PF_2 = 2a$, which says “I am $2a$ closer to $F_2$ than to $F_1$”. That condition describes only the points on the side of $F_2$: one branch. The equation $PF_1 - PF_2 = -2a$ describes the mirror-image branch on the side of $F_1$. The absolute value accepts both, so a hyperbola is one curve made of two separate pieces.

How large can the constant be

Not every constant produces a curve. Let $2c$ be the distance $F_1F_2$ between the foci, and pick a point $P$ of the set that is off the line through the foci, so $P$, $F_1$, $F_2$ form a real triangle. In any triangle, one side is always shorter than the sum of the other two, and that squeeze also bounds how far apart two sides can be: the difference of two sides is always strictly less than the third side. Applied here, that is $|PF_1 - PF_2| < F_1F_2$, so the constant must satisfy $2a < 2c$, that is

The excluded cases are worth naming, because they tell you what a hyperbola is not. If $2a = 2c$ the only points that qualify lie on the line through the foci, outside the segment joining them, and you get two rays instead of a curve. If $2a > 2c$ nothing qualifies at all and the set is empty. A real hyperbola needs $0 < a < c$, and this is the exact opposite of the ellipse, where the sum condition forces $a > c$.

From the condition to the equation

Set up coordinates so the foci are $F_1(-c, 0)$ and $F_2(c, 0)$, one on each side of the origin. The origin is then the center of the hyperbola.

The sign flip is the whole story of this chapter, so put the two results side by side and stare at them. For an ellipse the sum condition forces $a > c$, and $b^2 = a^2 - c^2$. For a hyperbola the difference condition forces $c > a$, and

For a hyperbola $c$ is the largest of the three, and it is the hypotenuse of a right triangle with legs $a$ and $b$. For an ellipse $a$ is the largest and it is the hypotenuse. Same three letters, same Pythagorean relation, different letter in the starring role.

Reading the standard form

With the center at the origin and the foci on the $x$-axis, the hyperbola is

and every feature can be read straight off it.

Vertices. Set $y = 0$: then $x^2 = a^2$, so $x = \pm a$. The two points $(\pm a, 0)$ are the vertices, the turning points where each branch is closest to the center. The segment joining them is the transverse axis, of length $2a$.

No $y$-intercepts. Set $x = 0$: the equation becomes $-\frac{y^2}{b^2} = 1$, which no real $y$ satisfies. The curve never crosses the $y$-axis. In fact $\frac{x^2}{a^2} = 1 + \frac{y^2}{b^2} \ge 1$ forces $|x| \ge a$, so the vertical strip $-a < x < a$ is completely empty. That gap in the middle is what separates the two branches.

Foci. They sit at $(\pm c, 0)$ where $c = \sqrt{a^2 + b^2}$, so each focus is beyond its vertex: $c > a$ always.

The conjugate axis. The segment from $(0, -b)$ to $(0, b)$ has length $2b$ and is called the conjugate axis, even though it touches no point of the curve. So what is $b$ good for? It sets the slopes of the two lines the branches chase, which is the next section.

The central rectangle. Draw the rectangle with corners $(\pm a, \pm b)$. Its half-width is $a$ and its half-height is $b$, so the distance from the center to any corner is $\sqrt{a^2 + b^2} = c$. Swing a compass from a corner down to the axis and you land exactly on a focus. This rectangle is the fastest way to sketch a hyperbola by hand.

The other orientation. Swap the roles of the variables and the branches open up and down:

Now $y = \pm a$ at the vertices $(0, \pm a)$, the foci are $(0, \pm c)$, and $b^2 = c^2 - a^2$ exactly as before. Which term is positive is what tells you the orientation, and $a^2$ is always the denominator under the positive term. Nothing anywhere forces $a > b$ for a hyperbola. The curve $\frac{y^2}{4} - \frac{x^2}{9} = 1$ opens up and down even though $9 > 4$, because the $y^2$ term is the positive one. If you learned “the bigger denominator tells you the axis” from ellipses, unlearn it here: it is simply false for hyperbolas.

Why the curve hugs two lines

Look at the upper half of the right branch and solve the standard equation for $y$:

Now set that beside the straight line $y = \frac{b}{a}x$, which is the same formula with the $-a^2$ thrown away. Since $x^2 - a^2 < x^2$ and both sides are nonnegative, $\sqrt{x^2 - a^2} < x$, so the curve runs strictly below the line. The vertical gap between them at a given $x$ is

A number minus a radical is exactly the shape that a radical conjugate flattens. Multiply and divide by $x + \sqrt{x^2 - a^2}$, which is positive for every $x \ge a$, so it is legal:

The difference of two square roots has turned into a difference of squares, which is the entire point of the conjugate trick. Multiplying by $\frac{b}{a}$ gives an exact formula for the gap:

This one identity settles everything, with no appeal to infinity anywhere.

The gap is never zero. The numerator $ab$ is a fixed positive number and the denominator is a positive real number for every $x \ge a$, so $\text{gap}(x) > 0$ at every single $x$. The curve stays strictly below the line forever; it never touches it and never crosses it.

The gap can be made as small as you please. Since $\sqrt{x^2 - a^2} \ge 0$, the denominator is at least $x$, so

That single inequality says the gap shrinks at least as fast as $\frac{ab}{x}$ does, and $\frac{ab}{x}$ can be pushed below any number you name just by making $x$ big enough. Challenge it to get under $0.001$: once $x$ is bigger than both $a$ and $\frac{ab}{0.001}$, the bound $\frac{ab}{x}$ is already under $0.001$, so the true gap is even smaller. However small a distance you challenge it to beat, going far enough out along the branch always beats it. That is the entire content of the claim that the curve “approaches” the line.

Here are the numbers for $\frac{x^2}{9} - \frac{y^2}{16} = 1$, where $a = 3$, $b = 4$, and so $ab = 12$:

The gap never reaches $0$, and it never rises above the bound $\frac{12}{x}$: the two are equal exactly at the vertex, where the radical vanishes, and the gap falls strictly below the bound at every row after that.

By symmetry, the equation contains only $x^2$ and $y^2$, so whatever happens in the first quadrant is mirrored in the other three. The lower half of the right branch chases $y = -\frac{b}{a}x$, and the left branch chases both lines on its side. Those two lines,

are the asymptotes of the hyperbola.

The shortcut, and why it works. To find the asymptotes, replace the $1$ on the right with $0$:

That is not a coincidence. The hyperbola itself factors the same way, into $\left(\frac{x}{a} - \frac{y}{b}\right)\left(\frac{x}{a} + \frac{y}{b}\right) = 1$, which says the two factors are reciprocals of each other. Far out along a branch one factor is large, so the other must be tiny, meaning the point very nearly satisfies one of the two linear equations. The vertical-gap computation above is the precise version of that idea.

For the vertical form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ the same replacement gives $\frac{y^2}{a^2} = \frac{x^2}{b^2}$, so the asymptotes are $y = \pm\frac{a}{b}x$. The slope is $\frac{a}{b}$ this time. Rather than memorizing which fraction goes with which picture, set the right side to $0$ and solve; it works for both.

Sketching by hand. The diagonals of the central rectangle have slope $\pm\frac{b}{a}$ and pass through the center, so the diagonals are the asymptotes. Draw the rectangle, extend its diagonals, mark the vertices, and sweep each branch from its vertex out along the diagonals. That is the whole sketching procedure.

Rectangular hyperbolas

When $a = b$ the central rectangle is a square, and the asymptotes are

whose slopes $1$ and $-1$ multiply to $-1$: the two asymptotes are perpendicular. Such a hyperbola is called rectangular (or equilateral), and its equation can be written $x^2 - y^2 = a^2$.

Every rectangular hyperbola has the same eccentricity, whatever its size, because

You have already met a rectangular hyperbola without being told. The graph of $y = \frac{1}{x}$ from the rational functions chapter, with the two coordinate axes as its asymptotes, is a rectangular hyperbola resting on its side. That curve is a turned copy of $x^2 - y^2 = 2$.

Hyperbolas with a shifted center

Moving the center from the origin to $(h, k)$ uses the same replacement rule as every other graph transformation: put $x - h$ where $x$ was and $y - k$ where $y$ was.

has center $(h, k)$, vertices $(h \pm a,\ k)$, foci $(h \pm c,\ k)$, and asymptotes

Everything is measured from the new center; nothing else changes. An equation of the form $Ax^2 + Cy^2 + Dx + Ey + F = 0$ with $A$ and $C$ opposite in sign is a hyperbola or, in one degenerate case, a pair of intersecting lines. To find out which, group the terms, complete the square, and divide. The degenerate case is exactly the one where the constant lands on $0$. Then you are left with $\frac{(x-h)^2}{a^2} = \frac{(y-k)^2}{b^2}$, a difference of squares equal to zero, which factors into the two asymptote lines and is not a hyperbola at all.

Eccentricity

The eccentricity of a hyperbola is the same ratio used for the other conics:

Because $c > a$ for every hyperbola, $e > 1$ always. That single inequality is what sorts these two conic sections: an ellipse has $0 \le e < 1$ (its $c$ is smaller than its $a$), and a hyperbola has $e > 1$. Push a conic’s eccentricity up past $1$ and the closed oval bursts open into two branches.

Eccentricity also controls the shape directly. Divide $b^2 = c^2 - a^2$ by $a^2$:

Read that carefully, because the ratio and the slope are not the same thing in both orientations. The identity $\frac{b}{a} = \sqrt{e^2 - 1}$ holds for every hyperbola. It is the asymptote slope for the horizontal form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, whose slopes are $\pm\frac{b}{a}$. For the vertical form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ the slopes are $\pm\frac{a}{b}$, the reciprocal, so there the slope is $\frac{1}{\sqrt{e^2 - 1}}$ instead. Worked example 2 is the check: $\frac{y^2}{9} - \frac{x^2}{16} = 1$ has $e = \frac{5}{3}$ and $\sqrt{e^2 - 1} = \frac{4}{3}$, while its asymptote slope is the reciprocal $\frac{3}{4}$.

Take a horizontal hyperbola and watch the shape follow $e$. When $e$ is just above $1$ the slope $\sqrt{e^2 - 1}$ is close to $0$, the asymptotes are nearly flat, and the branches are narrow and sharply pointed. As $e$ grows the branches flare open. A vertical hyperbola with the same $e$ is the same picture turned a quarter turn. That hyperbola’s asymptotes are nearly vertical rather than nearly flat, and its branches are just as narrow, because either way they hug their own transverse axis. A rectangular hyperbola has $a = b$, so $\sqrt{e^2 - 1} = 1$ and $e = \sqrt{2}$, matching what we found above, and that is the one case where the two orientations agree on the slope.

Eccentricity is not only a shape number. Substituting $e = \frac{c}{a}$ into the converse computation from the proof turns $PF_1 = \bigl|\tfrac{c}{a}x + a\bigr|$ and $PF_2 = \bigl|\tfrac{c}{a}x - a\bigr|$ into $PF_1 = |ex + a|$ and $PF_2 = |ex - a|$: exact formulas for both focal distances, written with $e$ in place of $c$.

Common mistakes