topic badge
Hong Kong
Stage 4 - Stage 5

Graphing hyperbolas with translations


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. 

Graph of a hyperbola centred at $\left(h,k\right)$(h,k)


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).

and remembering the values,

  • $a$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=c2a2

the table below summarizes the characteristics of a translated hyperbola in both orientations:

Orientation Horizontal Major Axis Vertical Major Axis
Standard form $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2(yk)2b2=1 $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(yk)2a2(xh)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 $\left(h-c,k\right)$(hc,k) $\left(h,k+c\right)$(h,k+c) and $\left(h,k-c\right)$(h,kc)
Vertices $\left(h+a,k\right)$(h+a,k) and $\left(k-a,k\right)$(ka,k) $\left(h,k+a\right)$(h,k+a) and $\left(h,k-a\right)$(h,ka)
Transverse axis $y=k$y=k $x=h$x=h
Conjugate axis $x=h$x=h $y=k$y=k
Asymptotes $y-k=\pm\frac{b}{a}(x-h)$yk=±ba(xh) $y-k=\pm\frac{a}{b}(x-h)$yk=±ab(xh)

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 graph of the hyperbola.


Worked Examples

The hyperbola $\frac{x^2}{25}-\frac{y^2}{100}=1$x225y2100=1 when translated $3$3 units to the right and $4$4 units up will be described by the equation $\frac{\left(x-3\right)^2}{25}-\frac{\left(y-4\right)^2}{100}=1$(x3)225(y4)2100=1. Find the centre, vertices, foci and asymptotes of the translated hyperbola. Then draw the graph of the hyperbola.

  • What is the centre of the hyperbola?
    Think: The equation of the hyperbola is in the form $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2(yk)2b2=1 and the centre will be $\left(h,k\right)$(h,k).
    Do: The centre can just be read off from the equation as $\left(3,4\right)$(3,4).
    Reflect: The values of $h$h and $k$k appear after a minus sign in the standard form of the equation. If the equation was in the form $\frac{\left(x+3\right)^2}{a^2}\ldots$(x+3)2a2 then $h$h would equal $-3$3.
  • What are the vertices of the hyperbola?
    Think: For a hyperbola centred at $\left(h,k\right)$(h,k) and oriented horizontally, the vertices will be $a$a units horizontally in both directions from the centre, $\left(h\pm a,k\right)$(h±a,k)
    Do: From the equation $a^2=25$a2=25 so $a=5$a=5. The vertices will be at $\left(3\pm5,4\right)$(3±5,4), or written separately, $\left(-2,4\right),\left(8,4\right)$(2,4),(8,4).
    Reflect: Notice that the translated centre, and the vertices all share the same $y$y-component. They will always lie on the transverse axis. 
Hyperbola centred at the origin and translated hyperbola with its vertices and centre marked
  • What are the equations of the asymptotes of the hyperbola?
    Think: For a hyperbola of the form above the asymptotes are in the form $y=\pm\frac{b}{a}\left(x-h\right)+k$y=±ba(xh)+k. We have found the values of $a$a$h$h and $k$k. We need to find $b$b and substitute in the values.
    Do: From the equation $b^2=100$b2=100 so $b=10$b=10. The asymptotes will be $y=\pm\frac{10}{5}\left(x-3\right)+4$y=±105(x3)+4, which simplifies to $y=\pm2\left(x-3\right)+4$y=±2(x3)+4
  • Draw the graph of the hyperbola.
    Think: The graph of the hyperbola is horizontally oriented, and will have a pair of asymptotes that intercept at the centre $\left(3,4\right)$(3,4).
    Do: The graph of the hyperbola is given below.
Graph of a hyperbola $\frac{\left(x-3\right)^2}{25}-\frac{\left(y-4\right)^2}{100}=1$(x3)225(y4)2100=1

Practice questions

Question 1

The graph of the hyperbola $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(yk)2a2(xh)2b2=1 is the same graph as the graph of $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$y2a2x2b2=1, except the center is at $\left(h,k\right)$(h,k) rather than at the origin.

Loading Graph...

  1. Given the graph of $\frac{y^2}{16}-\frac{x^2}{4}=1$y216x24=1, find the graph of $\frac{\left(y-5\right)^2}{16}-\frac{x^2}{4}=1$(y5)216x24=1.

    Loading Graph...


    Loading Graph...


    Loading Graph...


    Loading Graph...


Question 2

The graph of $\frac{x^2}{16}-\frac{y^2}{9}=1$x216y29=1 is given below. Consider the translated hyperbola described by the equation $\frac{x^2}{16}-\frac{\left(y-2\right)^2}{9}=1$x216(y2)29=1.

Loading Graph...

  1. What are the coordinates of the centre of the translated hyperbola?

  2. What are the coordinates of the vertices of the translated hyperbola?

  3. What are the coordinates of the foci of the translated hyperbola?

  4. What are the equations of the asymptotes of the translated hyperbola?

  5. Select the graph of the translated hyperbola, $\frac{x^2}{16}-\frac{\left(y-2\right)^2}{9}=1$x216(y2)