Question

A hyperbola has its center at (0,0)a vertex of (0,19) and an asymptote of y=19/16^x Find the equation that describes the hyperbola.

Answer 1

Hyperbola centered at the origin, vertex (0,19), asymptote y = (19/16) x

The axis is the y axis so this hyperbola has the form

$y^2/a^2 -x^2/b^2 = 1$

We have (0,19) on the hyperbola so a=19.  I'm going to take a wild guess that b=16 so our hyperbola has equation:

$\dfrac{y^2}{19^2} -\dfrac{x^2}{16^2} = 1$

How do we find the asymptote?  I don't remember, it's going to be approximately the line from the origin to some point with really big x and y.

$\dfrac{y^2}{19^2} = \dfrac{x^2}{16^2} + 1$

As x gets big the +1 will matter less and less, so the asymptote is

$\dfrac{y^2}{19^2} = \dfrac{x^2}{16^2}$

$(16y)^2 - (19x)^2 = 0$

$(16y - 19x)(16y + 19x) = 0$

$y= \pm \dfrac{19}{16} x$

Our guess was right!

Answer:  $\dfrac{y^2}{19^2} -\dfrac{x^2}{16^2} = 1$