Question 770449
With the center at (1,0) and a focus 5 units to the right, at (6,0), you know that the focal distance is {{{c=5}}} anf that the transverse axis is along the x-axis.
(You also know that the other focus will be 5 units to the left of the center, at (4,0), and that is useful if you need to sketch the hyperbola).
The equation of a hyperbola with a transverse axis parallel to the x-axis and center at (h,k) is
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}
In this case, since the center is (1,0), the equation is
{{{(x-1)^2/a^2-y^2/b^2=1}}}
We need to find {{{a}}} and {{{b}}}.
{{{a}}} is the distance from a vertex to the center, half of the length of the transverse axis.
If the transverse axis is 6, the vertices are 6 units apart,
and they are {{{a=6/2=3}}} units from the center.
(If you need to sketch the hyperbola, the vertices are 3 units to the right and 3 units to the left of the center, at (-2,0) and (4,0)).
The other number you need for the equation of the hyperbola is {{{b}}} such that
{{{c^2=a^2+b^2}}}.
So {{{5^2=3^2+b^2}}}-->{{{25=9+b^2}}}-->{{{b^2=25-9}}}-->{{{b^2=16}}}-->{{{b=4}}}
So the equation of the hyperbola in question is
{{{highlight((x-1)^2/3^2-y^2/4^2=1)}}} or {{{highlight((x-1)^2/9-y^2/16=1)}}}