Question 605623
EQUATION FOR THE HYPERBOLA:
A hyperbola with the equation {{{x^2/a^2-y^2/b^2=1}}}
has an eccentricity of
{{{e=sqrt(1+b^2/a^2)}}}
and a latus rectum of {{{2*l=2b^2/a}}}
So from {{{2b^2/a=64/3}}} we get {{{b^2/a=32/3}}}
We also have
{{{sqrt(1+b^2/a^2)=5/3}}} --> {{{1+b^2/a^2=25/9}}} --> {{{b^2/a^2=25/9-1}}} --> {{{b^2/a^2=16/9}}} --> {{{b/a=4/3}}}
Combining both equations:
{{{(b^2/a)/(b/a)=(32/3)/(4/3)}}} --> {{{highlight(b=8)}}}
and then
{{{b/(b/a)=8/(4/3)=8*3/4}}} --> {{{highlight(a=6)}}}
That gives us the equation:
{{{x^2/6^2-y^2/8^2=1}}} or {{{highlight(x^2/36-y^2/64=1)}}}
 
GRAPHING:
With {{{a=6}}} and {{{b=8}}}, we can draw that box that gives us the vertices and the asymptotes.
{{{drawing(300,300,-10,10,-10,10,
grid(1),
red(rectangle(-6,-8,6,8)),
blue(line(-9,-12,9,12)),
blue(line(-9,12,9,-12)),
blue(circle(6,0,0.4)),blue(circle(-6,0,0.4))
)}}}
We also know that
{{{e=c/a}}} so {{{5/3=c/6}}} --> {{{highlight(c=10)}}}
which tells us that the foci are at distance 10 from the center.
And since the latus rectum is 64/3, there are points of the hyperbola, 32/3 above and below the foci.
That gives us the location of the foci and 4 more points of the hyperbola
{{{drawing(300,300,-25,25,-25,25,
grid(1),
red(rectangle(-6,-8,6,8)),
blue(line(-21,-28,21,28)),
blue(line(-21,28,21,-28)),
blue(circle(6,0,1)),blue(circle(-6,0,1)),
green(circle(10,0,1)),green(circle(-10,0,1)),
blue(circle(10,32/3,1)),blue(circle(-10,32/3,1)),
blue(circle(10,-32/3,1)),blue(circle(-10,-32/3,1))
)}}} I have the foci (green circles) and six points of the hyperbola (blue circles marking the vertices and the ends of the latera recta).
Now, I would just connect the points of the hyperbola that I found with smooth curved arches ) ( that hug the asymptotes towards their ends.
{{{graph(300,300,-25,25,-25,25,4x/3,-4x/3,x^2/36-y^2/64>1)}}}