Question 1204150
The standard form you need is:

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

where: 

{{{a>b}}}

{{{2a}}}= the length of the major axis (vertical)

{{{2b}}}= the length of the minor axis (horizontal)

({{{h}}},{{{k}}}) = center of ellipse

you are given:

{{{(x-4)^2/9 + (y-3)^2/25}}}

Therefore:

{{{h=4}}} , {{{k=3}}}

{{{b^2 = 9}}}, {{{a^2=25}}}

{{{b = 3}}},{{{ a = 5}}}

C({{{h}}},{{{k}}}) = ({{{4}}},{{{3}}})


The coordinates of the endpoints of the major axis are: ({{{4}}},{{{3}}}±{{{5}}}) or ({{{4}}},{{{8}}}), ({{{4}}},{{{-2}}})

a. Maximum point on the major axis:({{{4}}},{{{8}}})
b. Minimum point on the major axis:({{{4}}},{{{-2}}})



The coordinates of the endpoints of the minor axis are: ({{{4}}}±{{{3}}},{{{3}}}) or ({{{7}}},{{{3}}}), ({{{1}}},{{{3}}})

c. Maximum point on the minor axis: ({{{7}}},{{{3}}})
d. Minimum point on the minor axis: ({{{1}}},{{{3}}})

e. Maximum focal point:
f. Minimum focal point:

Focal points are at the distance  {{{sqrt(25-9)= sqrt(16)= 4}}} from the center vertically, up and down

so, foci are
 ({{{4}}},{{{3+4}}})= ({{{4}}},{{{7}}}) above  center
and 
({{{4}}},{{{3-4}}})= ({{{4}}},{{{-1}}})  below center

 e. Maximum focal point: ({{{4}}},{{{7}}})
f. Minimum focal point: ({{{4}}},{{{-1}}})

 

{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(4,3,.12), locate(4,3,C(4,3)),
circle(4,8,.12), locate(4,8,max(4,8)),
circle(4,-2,.12), locate(4,-2,min(4,-2)),
circle(4,7,.12), locate(4,7,F(4,7)),
circle(4,-1,.12), locate(4,-1,F(4,-1)),
circle(7,3,.12), locate(7,3,max(7,3)),
circle(1,3,.12), locate(1,3,min(1,3)),
graph( 600, 600, -10, 10, -10, 10,(1/3)(9-5sqrt(-x^2+8x-7)),(1/3)(5sqrt(-x^2+8x-7)+9)) )}}}