Question 1185036
<pre>
We plot those given points and sketch in a rough sketch of the
hyperbola.  All "conic" graphs curve around their focal points.

{{{drawing(400,200,-12,12,-6,6,graph(400,200,-12,12,-6,6,(3/4)sqrt(-16+x^2)),circle(0,0,.3),
circle(5,0,.3),circle(-5,0,.3),circle(4,0,.3),circle(-4,0,.3),

graph(400,200,-12,12,-6,6,(-3/4)sqrt(-16+x^2)) )}}}

The equation of all hyperbolas that open right and left have equations,

{{{(x-h)^2/a^2}}}{{{""-""}}}{{{(y-k)^2/b^2}}}{{{""=""}}}{{{1}}}

where (h,k) is the center, "a" is the distance between the center and
either vertex. "c" is the distance between the center and either focal
point (or focus).  But "c" does not appear in the standard equation of
a hyperbola. We calculate "b<sup>2</sup>" by the Pythagorean theorem:

{{{c^2=a^2+b^2}}}
{{{5^2=4^2+b^2}}}
{{{25=16+b^2}}}
{{{9=b^2}}}

The center is (0,0), so h=0, k=0, a<sup>2</sup>=4^2=16, b<sup>2</sup>=3^2=9.

Substitute in

{{{(x-h)^2/a^2}}}{{{""-""}}}{{{(y-k)^2/b^2}}}{{{""=""}}}{{{1}}}

{{{(x-0)^2/16^""}}}{{{""-""}}}{{{(y-0)^2/9^""}}}{{{""=""}}}{{{1}}}

{{{x^2/16^""}}}{{{""-""}}}{{{y^2/9^""}}}{{{""=""}}}{{{1}}}

Edwin</pre>