Question 1194603
<pre>
{{{y^2/36^""+x^2/25^""=1}}}

Write as

{{{(y-0)^2/6^2+(x-0)^2/5^2=1}}}

Compare to 

{{{(y-k)^2/a^2+(x-h)^2/b^2=1}}} which is an ellipse taller than it is wide

h=0, k=0, a=6, b=5 (a is larger than b in an ellipse)

From the center (h,k) = (0,0), go a=6 units up and down, to form
the major axis, and go b=5 units right and left, to form the minor axis.

The vertices are (h,k±a) = (0,a) and (0,-6)

The co-vertices are (h±a,k) = (-5,0) and (0,-5)

The major and minor axes are in green below:

{{{drawing(350,400,-7,7,-8,8, graph(350,400,-7,7,-8,8),

green(line(-5,0,5,0), line(0,-6,0,6))  )}}}

We sketch in the ellipse:

{{{drawing(350,400,-7,7,-8,8, graph(350,400,-7,7,-8,8),
arc(0,0,10,-12),
green(line(-5,0,5,0), line(0,-6,0,6))  )}}}

We calculate c, the distance from Center to foCus.

{{{c^2=a^2-b^2}}}
{{{c^2=6^2-5^2}}}
{{{c^2=36-25}}}
{{{c^2=11}}}
{{{c=sqrt(11)}}}

So the foci are (h,k±c) = {{{(matrix(1,3,0,",",sqrt(11)))}}} and {{{(matrix(1,3,0,",",-sqrt(11)))}}}

Draw the chords through them (in blue), which are 
the latus recta:

{{{drawing(350,400,-7,7,-8,8, graph(350,400,-7,7,-8,8),
arc(0,0,10,-12),
blue(line(-25/6,sqrt(11),25/6,sqrt(11))),
blue(line(-25/6,-sqrt(11),25/6,-sqrt(11))),
green(line(-5,0,5,0), line(0,-6,0,6))  )}}}

We find the ends of the latus recta by substituting y=±√11 for y
in the equation of the ellipse and solving for x

{{{("" +- sqrt(11))^2/36^""+x^2/25^""=1}}}
{{{11^""/36^""+x^2/25^""=1}}}
{{{x^2/25^""=1-11^""/36^""}}}
{{{x^2/25^""=25^""/36^""}}}
{{{x^2=25^2/36^""}}}
{{{x="" +- 5/6}}}

So the ends of the latus recta are 

{{{(matrix(1,3,5/6,",",sqrt(11)))}}},{{{(matrix(1,3,5/6,",",-sqrt(11)))}}},
{{{(matrix(1,3,-5/6,",",sqrt(11)))}}},{{{(matrix(1,3,-5/6,",",-sqrt(11)))}}}

Edwin</pre>