Hyperbola


A hyperbola is the set of all the points in the plane such that the diference of the distances of each point to the two focal points is constant.


Test the definition yourself, by moving the point on the hyperbola, notice how the sum of the distances is always 66.




There are two types of hyperbolas, horizontals and verticals.


Switch to vertical

The formula of a horizontal hyperbola with vertex at the origin (0,0)(0,0) is

x2a2y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

where aa and bb will determine the shape of the hyperbola and of its asymptotes.


For example, the following hyperbola has formula:

x2(2)2+y2(1)2=1,\frac{x^2}{(2)^2} + \frac{y^2}{(1)^2} = 1,

and asymptotes y=±12xy=\pm\frac{1}{2}x





Switch to vertical


Horizontal hyperbola with vertex at (0,0)(0,0)

  • Focal points: (±c,0)(\pm c,0).
  • Asymptotes: y=±baxy=\pm\frac{b}{a}x.

Where cc is calculated as c=a2+b2.c = \sqrt{a^2+b^2}.

Let's see how the focal points affect the shape of the hyperbola for a fixed aa.


And what happens if we change the center?

If we move the hyperbola horizontally or vertically, we need to aply the corresponding transformations to the variables xx and yy.

  • Horizontal shift of hh units: Change the value of xx for xhx-h.

  • Vertical shift of kk units: Change the value of yy for yky-k.

Together with the variables xx and yy, the other elements affected by the translation are the focus points, these are now (±c+h,k)(\pm c + h, k) for horizontal hyperbolas and (h,±c+k)(h, \pm c +k) for vertical hyperbolas.


Perfect! Now let's perform this transformations on the formula of the hyperbola.


Switch to vertical

The formula of a horizontal hyperbola with vertex in (h,k)(h,k) is:

(xh)2a2(yk)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1

let's see what this looks like. The fomula:

(x(0))2a2(y(0))2b2=1,\frac{\big(x-(0)\big)^2}{a^2} - \frac{\big(y-(0)\big)^2}{b^2} = 1,

gives us the following hyperbola:




To sum up:


Switch to vertical


We've learnt how to plot a general horizontal hyperbola with:

  • vertex at (h,k)=(0,0)(h,k) = (0, 0)
  • horizontal and vertical axis lengths a=2a = 2 and b=1b = 1
  • focus point (h+c,k)=(0+1.41,0)(h+c,k) = (0 + 1.41,0)
  • asymptote y(0)=±11(x(0))y-(0) = \pm\frac{1}{1}(x-(0))
(x(0))2(1)2(y(0))2(1)2,\frac{(x-(0))^2}{(1)^2} - \frac{(y-(0))^2}{(1)^2},