Question 1185041
<br>
The vertices are (-3,-3) and (5,-3).  That means<br>
(1) the center is (1,-3)
(2) the branches open right and left<br>
The standard form of the equation is then<br>
{{{(x-1)^2/a^2-(y+3)^2/b^2=1}}}<br>
The slope of one asymptote, 1/2, is equal to b/a.<br>
{{{b/a=1/2}}}
{{{a=2b}}}<br>
Substitute that into the equation:<br>
{{{(x-1)^2/(4b^2)-(y+3)^2/b^2=1}}}<br>
Substitute the known point (5,-3) into the equation to find b:<br>
{{{4^2/(4b^2)-0=1}}}
{{{16/(4b^2)=1}}}
{{{b^2=4}}}
{{{b=2}}}<br>
Then<br>
{{{a=2b=4}}}<br>
ANSWER:
{{{(x-1)^2/16-(y+3)^2/4=1}}}<br>