Question

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

Answer 1

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]}