Ikleyn gets all bent out of shape if you don't have 
everything exactly the way she wants it.

It's obvious you were asking about a hyperbola.

Plot the two foci:



The center is halfway between them at (2,1).  Plot the center.



The center is the midpoint of the transverse axis, which is 4
units.  So draw it half of that, 2 units left and right of the
center.  The ends of the transverse axes are the vertices of the
hyperbola.



a = semi-transverse axis = half of the transverse axis = 2
c = the distance between either focus and the center. That's 5
units from focus (7,1) and center (2,1)

The Pythagorean relationship for all hyperbolas is c2=a2+b2





Now we have enough to write the equation of the hyperbola,
with center (h,k) = (2,1)





b is the length of the semi-conjugate axis. That's about 4.6
The center is the midpoint of the conjugate axis, so,
draw the conjugate axis, from 4.6 units below the center to
4.6 units above the center:


 

Now draw the defining rectangle which has the ends of the
transverse and conjugate axes as midpoints.



Extend the graph and draw the asymptotes, which are the
extended diagonals of the defining rectangle:



Sketch in the hyperbola approaching those asymptotes:



Edwin