Find the equation of a hyperbola
We now examine the method of finding the equation of a hyperbola given certain identifying information.
Central hyperbolas
The standard form for a central hyperbola depends on the orientation of the hyperbola.
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| The foci, vertices and asymptotes of a horizontal hyperbola. | The foci, vertices and asymptotes of a vertical hyperbola. |
The equations and attributes can be summarized in the table below, given the following:
- $a$a is the distance from the center to a vertex.
- $c$c is the distance from the center to a focus.
- $b$b can be found using the relationship $b^2=c^2-a^2$b2=c2−a2.
| Orientation | Horizontal Transverse Axis | Vertical Transverse Axis |
|---|---|---|
| Standard form | $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2−y2b2=1 | $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$y2a2−x2b2=1 |
| Center | $\left(0,0\right)$(0,0) | $\left(0,0\right)$(0,0) |
| Foci | $\left(c,0\right)$(c,0) and $\left(-c,0\right)$(−c,0) | $\left(0,c\right)$(0,c) and $\left(0,-c\right)$(0,−c) |
| Vertices | $\left(a,0\right)$(a,0) and $\left(-a,0\right)$(−a,0) | $\left(0,a\right)$(0,a) and $\left(0,-a\right)$(0,−a) |
| Transverse axis | $y=0$y=0 | $x=0$x=0 |
| Conjugate axis | $x=0$x=0 | $y=0$y=0 |
| Asymptotes | $y=\pm\frac{b}{a}x$y=±bax | $y=\pm\frac{a}{b}x$y=±abx |
Notice that in this form, it is always true that the $a$a value is first in the equation.
Translated Hyperbolas
If a hyperbola is translated horizontally or vertically from the center, the parameters $a$a, $b$b, and $c$c still have the same meaning. However, we must take into account that the center of the hyperbola has moved.
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| Centre of a translated hyperbola. |
Given the following definitions for $h$h and $k$k,
- $h$h denotes the translation in the horizontal direction from $\left(0,0\right)$(0,0).
- $k$k denotes the translation in the vertical direction from $\left(0,0\right)$(0,0).
the table below summarizes the standard form of a hyperbola in both orientations:
| Orientation | Horizontal Major Axis | Vertical Major Axis |
|---|---|---|
| Standard form | $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2−(y−k)2b2=1 | $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(y−k)2a2−(x−h)2b2=1 |
| Center | $\left(h,k\right)$(h,k) | $\left(h,k\right)$(h,k) |
| Foci | $\left(h+c,k\right)$(h+c,k) and $\left(h-c,k\right)$(h−c,k) | $\left(h,k+c\right)$(h,k+c) and $\left(h,k-c\right)$(h,k−c) |
| Vertices | $\left(h+a,k\right)$(h+a,k) and $\left(k-a,k\right)$(k−a,k) | $\left(h,k+a\right)$(h,k+a) and $\left(0,-a\right)$(0,−a) |
| Transverse axis | $y=k$y=k | $x=h$x=h |
| Conjugate axis | $x=h$x=h | $y=k$y=k |
| Asymptotes | $y-k=\pm\frac{b}{a}(x-h)$y−k=±ba(x−h) | $y-k=\pm\frac{a}{b}(x-h)$y−k=±ab(x−h) |
Essentially, the information is the same as the central hyperbola. But the values of $h$h and $k$k are added to the $x$x and $y$y-values (respectively) for each characteristic.
Once we establish certain information about the hyperbola, we can use the relationships summarized in the tables to determine the equation.
Worked Examples
Example 1
Find the equation of the hyperbola shown below.
Think: The centre is at $\left(0,0\right)$(0,0) and it is aligned horizontally so it will have the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2−y2b2=1. The vertices will be in the form $\left(\pm a,0\right)$(±a,0) and the foci will be at $\left(\pm c,0\right)$(±c,0) and we can find $b$b using the equation $c^2=a^2+b^2$c2=a2+b2.
Do: From the information above, and determining that a vertex is at $\left(5,0\right)$(5,0) and a focus is at $\left(13,0\right)$(13,0) we know that $a=5$a=5 and $c=13$c=13. We can then use the equation above:
| $c^2$c2 | $=$= | $a^2+b^2$a2+b2 | (The equation relating $a$a, $b$b and $c$c) |
| $b^2$b2 | $=$= | $c^2-a^2$c2−a2 | (Rearranging the equation to make $b^2$b2 the subject) |
| $b$b | $=$= | $\sqrt{c^2-a^2}$√c2−a2 | (Rearranging the equation to make $b$b the subject) |
| $b$b | $=$= | $\sqrt{13^2-5^2}$√132−52 | (Substituting the values of $a$a and $b$b) |
| $b$b | $=$= | $\sqrt{144}$√144 | (Evaluating the numbers inside the square root) |
| $b$b | $=$= | $12$12 | (Evaluating the square root) |
So the equation of the hyperbola will be $\frac{x^2}{5^2}-\frac{y^2}{12^2}=1$x252−y2122=1 or if we evaluate the denominators, $\frac{x^2}{25}-\frac{y^2}{144}=1$x225−y2144=1.
Example 2
Find the equation of the hyperbola that passes through the point $\left(12,5\right)$(12,5) and has a vertices at $\left(0,\pm3\right)$(0,±3).
Think: First to determine the orientation of the hyperbola we can look at the vertices. The vertices will lie on $y$y-axis as both $x$x-coordinates are the same. The vertices also tell us the value of $a$a. Using the equation and the point on the hyperbola we can determine the value of $b$b.
Do: The value of $a$a will be $3$3. The equation of the hyperbola will be $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$y2a2−x2b2=1. If we substitute the value of $a$a and the point's $x$x and $y$y-coordinates for $x$x and $y$y in the equation.
| $\frac{y^2}{a^2}-\frac{x^2}{b^2}$y2a2−x2b2 | $=$= | $1$1 | (Equation of hyperbola) |
| $\frac{x^2}{b^2}$x2b2 | $=$= | $\frac{y^2}{a^2}-1$y2a2−1 | (Making the term with $b$b the subject) |
| $\frac{x^2}{b^2}$x2b2 | $=$= | $\frac{y^2-a^2}{a^2}$y2−a2a2 | (Combining two terms) |
| $b^2$b2 | $=$= | $\frac{x^2a^2}{y^2-a^2}$x2a2y2−a2 | (Making $b^2$b2 the subject) |
| $b$b | $=$= | $\frac{xa}{\sqrt{y^2-a^2}}$xa√y2−a2 | (Taking the square root of both sides) |
| $b$b | $=$= | $\frac{12\times3}{\sqrt{5^2-3^2}}$12×3√52−32 | (Substituting) |
| $b$b | $=$= | $\frac{36}{4}$364 | (Simplifying the numerator and denominator) |
| $b$b | $=$= | $9$9 | (Simplifying the fraction) |
So the equation of the hyperbola will be $\frac{y^2}{3^2}-\frac{x^2}{9^2}=1$y232−x292=1 or if we evaluate the denominators $\frac{y^2}{9}-\frac{x^2}{81}=1$y29−x281=1.
Example 3
Find the equation of the hyperbola with the asymptotes $y=\pm2\left(x-3\right)+4$y=±2(x−3)+4, and vertices at $\left(-2,4\right)$(−2,4) and $\left(8,4\right)$(8,4).
Think: From the asymptotes and vertices we can tell that this hyperbola is not centred at the origin. From the vertices we can also tell that the hyperbola is oriented horizontally so its asymptotes will be in the form $y=\pm\frac{b}{a}\left(x-h\right)+k$y=±ba(x−h)+k. From this we can read off $h$h and $k$k. The distance between the two vertices will be $2a$2a. We can use the gradient of the asymptote to determine $b$b.
Do: The value of $h=3$h=3 and $k=4$k=4. The distance between the two vertices will be $2a=\left|8-\left(-2\right)\right|$2a=|8−(−2)| so $a=5$a=5.
Now the gradient $\frac{b}{a}$ba should be equal to $2$2. So we can use this and the value of $a$a to find $b$b.
| $\frac{b}{a}$ba | $=$= | $2$2 |
| $\frac{b}{5}$b5 | $=$= | $2$2 |
| $b$b | $=$= | $2\times5$2×5 |
| $b$b | $=$= | $10$10 |
So the equation of the hyperbola will be $\frac{\left(x-3\right)^2}{5^2}-\frac{\left(y-4\right)^2}{10^2}=1$(x−3)252−(y−4)2102=1 or if we evaluate the denominators, $\frac{\left(x-3\right)^2}{25}-\frac{\left(y-4\right)^2}{100}=1$(x−3)225−(y−4)2100=1.
Reflect: Notice that we also could have found the values of $h$h and $k$k using the vertices. The coordinate in the middle of the two points will be the centre. So in this case the point between the vertices will be $\left(\frac{-2+8}{2},4\right)$(−2+82,4) which simplifies to $\left(3,4\right)$(3,4). These are values we found for $h$h and $k$k.
Practice questions
Question 1
Consider the hyperbola with vertices at $\left(-11,1\right)$(−11,1) and $\left(1,1\right)$(1,1), and asymptotes $y=\pm\frac{2}{3}\left(x+5\right)+1$y=±23(x+5)+1.
In what form will the equation of the hyperbola be?
$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2−(y−k)2b2=1
A$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(y−k)2a2−(x−h)2b2=1
BWhat are the coordinates of the centre of the hyperbola?
What is distance from the vertex to the centre, $a$a?
What is the value of $b$b?
Write the equation of the hyperbola, in the form found in part (a).
Question 2
Consider the hyperbola with vertices $\left(\pm4,0\right)$(±4,0) that passes through the point $\left(5,-6\right)$(5,−6).
In what form will the equation of the hyperbola be?
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2−y2b2=1
A$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$y2a2−x2b2=1
BWhat is the value of $b$b?
State the equation of the hyperbola in the form from part (a).
Question 3
Find the equation, in standard form, of the hyperbola with vertices $\left(1,-11\right)$(1,−11) and $\left(1,1\right)$(1,1), and asymptotes $y=\pm\frac{2}{3}\left(x-1\right)$y=±23(x−1) $-$− $5$5.
In what form will the equation of the hyperbola be?
$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2−(y−k)2b2=1
A$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(y−k)2a2−(x−h)2b2=1
BWrite the equation of the hyperbola, in the form found in part (a).