Question 1079298
<pre><b>
There are two cases.

Case 1.  If the hyperbola has equation {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}},

then for p < a,  any ellipse whose equation is{{{(x-h)^2/a^2+(y-k)^2/p^2=1}}}
will share the vertices (h+a,k) and (h-a,k) and the center (h,k).

Case 2.  If the hyperbola has equation {{{(y-k)^2/a^2-(x-h)^2/b^2=1}}},

then for p < a,  any ellipse whose equation is{{{(y-k)^2/a^2+(x-h)^2/p^2=1}}}
will share the vertices (h,k+a) and (h,k-a) and the center (h,k). 

Example and graph for case 1.

{{{(x-2)^2/3^2-(y+3)^2/4^2=1}}} <-- hyperbola
{{{(x-2)^2/3^2+(y+3)^2/2^2=1}}} <-- ellipse

{{{drawing(400,375,-6,10,-10,5,graph(400,375,-6,10,-10,5,(4/3)sqrt(x^2-4x-5)-3),
green(line(-1,-3,5,-3)),
graph(400,375,-6,10,-10,5,-(4/3)sqrt(x^2-4x-5)-3),
arc(2,-3,6,-4),circle(2,-3,.1),locate(-3.3,-3+.3,"(-1,-3)"),
locate(5,-3+.3,"(5,-3)"),circle(-1,-3,.1), circle(5,-3,.1),
locate(2-.5,-3.1,"(2,-3)"),
blue(rectangle(-1,-7,5,1),line(-1,-7,5,-7),line(-1,1,5,1),line(-10,-19,11,9),
line(-10,13,11,-15)) 
 

   )}}}

Example and graph for case 2.

{{{(y-3)^2/5^2-(x-2)^2/2^2=1}}} <-- hyperbola
{{{(y-3)^2/5^2+(x-2)^2/2^2=1}}} <-- ellipse

{{{drawing(4000/13,400,-8,12,-10,16,graph(4000/13,400,-8,12,-10,16,(5/4)sqrt(x^2-4x+20)+3),
green(line(2,8,2,-2)),
graph(4000/13,400,-8,12,-10,16,-(5/4)sqrt(x^2-4x+20)+3),
arc(2,3,8,-10),circle(2,3,.1),locate(2.2,3.5,"(2,3)"),

locate(.59,-2.3,"(2,-2)"),locate(.59,9.3,"(2,8)"),
blue(rectangle(-2,-2,6,8),line(-2,-2,6,-2),line(-2,8,6,8),line(-14,-17,14,18),
line(-14,23,14,-12)) 
 

   )}}}

Edwin</pre></b>