How do you write the equation for a hyperbola given vertices (0,-4) and (0,4), conjugate axis of lengths 14 units?

1 Answer
Jun 30, 2017

Answer:

Equation of hyperbola is #y^2/16-x^2/49=1#

Explanation:

As the vertices are #(0,-4)# and #(0,4)# i.e. along #y#-axis, we have a vertical hyperbola and equation of hyperbola is of the type

#(y-k)^2/a^2-(x-h)^2/b^2=1#, where #(h,k)# is center of hyperbola and #b>a#

As both vertices are equidistant from origin, center of hyperbola is #(0,0)# and the equation is

#y^2/4^2-x^2/b^2=1#

Further as length of conjugate axis is #14#, we have #b=14/2-7#

and hence equation of hyperbola is #y^2/16-x^2/49=1#
graph{y^2/16-x^2/49=1 [-20, 20, -10, 10]}