Question 83900
The given equation is:  {{{((x-3)^2/25) + ((y-4)^2/16) = 1}}} 


Comparing this to the standard equation, we get:  


{{{((x - h)^2/a^2) + ((y - k)^2/b^2) = 1}}} 


We find that the centre of the ellipse given by (h,k) is equal to (3, 4) 


The foci of the Ellipse is given by (+- ae, 0) 


So we first find the eccentricity using the formula: 


e =  {{{ sqrt(1 - (b^2/a^2)) }}} 


e = {{{ sqrt(1 - 16/25)) }}} 


e = 3/5 


So now focus is:  (+- ae, 0)  = (+ -3, 0) 


Now the vertex is (+ - a, 0)  = (+-  5, 0) 


hence, the solution