Lesson 11.4: Hyperbolas

New Concept

A hyperbola is a unique conic section defined by the difference in distances from two fixed points. You'll learn to graph its distinct two-branched shape from its equation, whether it's centered at the origin or elsewhere.

What’s next

Next, you'll master graphing these curves through interactive examples and practice cards, starting with hyperbolas centered at the origin.

Property

A hyperbola is all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a focus of the hyperbola.
The line through the foci is called the transverse axis. The two points where the transverse axis intersects the hyperbola are each a vertex of the hyperbola. The midpoint of the segment joining the foci is called the center of the hyperbola. The line perpendicular to the transverse axis that passes through the center is called the conjugate axis. Each piece of the graph is called a branch of the hyperbola.

Examples

• A point on a hyperbola is 12 units from one focus and 5 units from the other. The constant difference for this hyperbola is $|12 - 5| = 7$.
• The line segment connecting a hyperbola's two vertices is part of the transverse axis, and its midpoint is the center of the hyperbola.
• A hyperbola consists of two separate, mirror-image curves called branches. These branches open infinitely, approaching but never touching the asymptotes.

Explanation

A hyperbola has two U-shaped curves called branches that open away from each other. It's defined by the constant difference in distances from any point on the curve to two fixed points (foci), unlike an ellipse, which uses a constant sum.

Property

The standard form of the equation of a hyperbola with center $(0, 0)$ is

For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the transverse axis is on the x-axis, it opens left and right, vertices are $(\pm a, 0)$, and asymptotes are $y = \pm \frac{b}{a}x$. For $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, the transverse axis is on the y-axis, it opens up and down, vertices are $(0, \pm a)$, and asymptotes are $y = \pm \frac{a}{b}x$.
To graph: Write the equation in standard form. Find vertices from the positive term's denominator, $a^2$. Sketch a rectangle using points $(\pm a, 0)$ and $(0, \pm b)$ (or vice versa depending on orientation). Draw asymptotes through the rectangle's diagonals. Draw the branches from the vertices, approaching the asymptotes.

Examples

• To graph $\frac{x^2}{16} - \frac{y^2}{9} = 1$, note the center is $(0,0)$ and it opens horizontally since $x^2$ is positive. The vertices are $(\pm 4, 0)$. The guide rectangle uses $(0, \pm 3)$, giving asymptotes $y = \pm \frac{3}{4}x$.
• To graph $\frac{y^2}{64} - \frac{x^2}{25} = 1$, note the center is $(0,0)$ and it opens vertically since $y^2$ is positive. The vertices are $(0, \pm 8)$. The guide rectangle uses $(\pm 5, 0)$, giving asymptotes $y = \pm \frac{8}{5}x$.
• To graph $4y^2 - 25x^2 = 100$, first divide by 100 to get $\frac{y^2}{25} - \frac{x^2}{4} = 1$. This is a vertical hyperbola with vertices at $(0, \pm 5)$ and asymptotes $y = \pm \frac{5}{2}x$.

Explanation

This equation describes a hyperbola centered at the origin. The term with the positive coefficient ($x^2$ or $y^2$) tells you if it opens horizontally or vertically. The values $a$ and $b$ create a guide rectangle for sketching the asymptotes.

Property

The standard form of the equation of a hyperbola with center $(h, k)$ is:

If the $(x-h)^2$ term is positive, the transverse axis is horizontal. If the $(y-k)^2$ term is positive, the transverse axis is vertical. The vertices are $a$ units from the center along the transverse axis.
To graph: Identify the center $(h,k)$, $a$, and $b$. Determine orientation. Sketch a rectangle centered at $(h,k)$ that is $2a$ units wide (or tall) and $2b$ units tall (or wide). Draw asymptotes through the diagonals and draw the branches from the vertices.

Examples

• The graph of $\frac{(x-3)^2}{25} - \frac{(y-1)^2}{4} = 1$ is a horizontal hyperbola centered at $(3,1)$. The vertices are $5$ units left and right of the center, at $(-2,1)$ and $(8,1)$.
• The graph of $\frac{(y+4)^2}{9} - \frac{(x+2)^2}{16} = 1$ is a vertical hyperbola centered at $(-2,-4)$. The vertices are $3$ units up and down from the center, at $(-2,-1)$ and $(-2,-7)$.
• The equation $\frac{(x-5)^2}{49} - (y+3)^2 = 1$ can be written as $\frac{(x-5)^2}{49} - \frac{(y+3)^2}{1} = 1$. It is a horizontal hyperbola centered at $(5, -3)$ with $a=7$ and $b=1$.

Explanation

This form describes a hyperbola whose center is shifted from the origin to a new point $(h, k)$. The values $h$ and $k$ indicate the horizontal and vertical shifts. All other properties, like orientation and shape, are relative to this new center.

Property

To write the equation of a hyperbola in standard form from its general form, you must complete the square. First, group the x-terms and y-terms together and move the constant to the other side of the equation. Factor out the leading coefficient from each variable's group. Complete the square for each binomial, being sure to add the correct corresponding value to the other side. Finally, divide both sides by the new constant to make the equation equal to 1.

Examples

• Convert $x^2 - 4y^2 - 2x - 24y - 39 = 0$. Group terms: $(x^2 - 2x) - 4(y^2 + 6y) = 39$. Complete the square: $(x-1)^2 - 4(y+3)^2 = 39 + 1 - 36 = 4$. Divide by 4: $\frac{(x-1)^2}{4} - \frac{(y+3)^2}{1} = 1$.
• Convert $9y^2 - 16x^2 + 90y + 32x + 65 = 0$. Group terms: $9(y^2 + 10y) - 16(x^2 - 2x) = -65$. Complete the square: $9(y+5)^2 - 16(x-1)^2 = -65 + 225 - 16 = 144$. Divide by 144: $\frac{(y+5)^2}{16} - \frac{(x-1)^2}{9} = 1$.
• Convert $x^2 - y^2 + 6x + 4y + 4 = 0$. Group terms: $(x^2 + 6x) - (y^2 - 4y) = -4$. Complete the square: $(x+3)^2 - (y-2)^2 = -4 + 9 - 4 = 1$. The standard form is $(x+3)^2 - (y-2)^2 = 1$.

Explanation

Completing the square transforms a hyperbola's general equation into the much more useful standard form. This process reveals the hyperbola's center $(h, k)$, its orientation, and the values of $a$ and $b$ needed for graphing.

Property

To identify a conic from its general equation $Ax^2 + By^2 + Cx + Dy + E = 0$, examine the squared terms:

• Parabola: Only one variable is squared (either an $x^2$ term or a $y^2$ term, but not both).
• Circle: The $x^2$ and $y^2$ terms have the same coefficients (A = B).
• Ellipse: The $x^2$ and $y^2$ terms have the same sign but different coefficients (A and B are both positive or both negative, and A $\neq$ B).
• Hyperbola: The $x^2$ and $y^2$ terms have different signs (one is positive, one is negative).

Examples

• In $3x^2 - 5y^2 + 6x - 20y - 47 = 0$, the $x^2$ and $y^2$ terms have opposite signs, so it is a hyperbola.
• In $y = 4x^2 - 8x + 7$, only the $x$ variable is squared, so it is a parabola.
• In $2x^2 + 5y^2 - 12x + 10y - 3 = 0$, the $x^2$ and $y^2$ terms have the same sign (both positive) but different coefficients (2 and 5), so it is an ellipse.

Explanation

By inspecting the coefficients and signs of the $x^2$ and $y^2$ terms in a conic's general equation, you can quickly determine its shape without graphing. This is the first step in deciding how to analyze or graph the equation.