Question 1199474
<pre>
I'll do this one instead.  It's done exactly the same as yours step by step:

{{{25x^2-16y^2-200x+96y=144}}}

Rearrange to get the terms in x together and the terms in y together.

{{{25x^2-200x-16y^2+96y=144}}}

Factor 25 out of the 1st and 2nd terms on the left, and factor -16 out of the
3rd and 4th terms on the left:

{{{25(x^2-8x)-16(y^2+6y)=144}}}

Skip a big space at the end of each parentheses because we're going to insert
a couple of terms in each one to complete the square.

{{{25(x^2-8x+"_______")-16(y^2+6y+"_______")=144}}}

In the first parentheses:
Multiply the coefficient of x, which is -8, by 1/2 getting -4.
Square -4, getting (-4)<sup>2</sup> or +16, so we add and subtract 16 in the
first blank, that is, put +16-16 in the first blank:

{{{25(x^2-8x+16-16)-16(y^2+6y+"_______")=144}}}

In the second parentheses:
Multiply the coefficient of y, which is -6, by 1/2 getting -3.
Square -3, getting (-3)<sup>2</sup> or +9, so we add and subtract 9 in the
second blank, that is, put +9-9 in the second blank:

{{{25(x^2-8x+16-16)-16(y^2+6y+9-9)=144}}}

Factor the first three terms in each parentheses:
            {{{x^2-8x+16=(x-4)(x-4)=(x-4)^2}}}
            {{{y^2-6y+9=(x-3)(x-3)=(x-3)^2}}}    
 
{{{25((x-4)^2-16)-16((x-3)^2-9)=144}}}

Distribute the 25 and the -16, being careful to leave the squared parentheses intact:

{{{25(x-4)^2-400-16(x-3)^2+144=144}}}

{{{25(x-4)^2-16(x-3)^2-256=144}}}

{{{25(x-4)^2-16(x-3)^2=400}}}

Divide every term by 400 to get 1 on the right side:

{{{25(x-4)^2/400-16(x-3)^2/400=400/400}}}

cancel:

{{{(x-4)^2/16-(x-3)^2/25=1}}}

Compare to

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

Where the center is (h,k) = (4,3),
semi-transverse axis length = a = 4, 
semi-conjugate axis length = b = 5

Plot the center (h,k) = (4,3).
Draw the transverse axis horizontally from the center a=4 units 
right and a=4 units left.

Draw the conjugate axis vertically from the center b=5 units 
upward and b=5 units downward.

{{{drawing(320,420,-4,12,-9,15, 

graph(320,420,-4,12,-9,15),

blue(line(0,3,8,3), line(4,-2,4,8)) 

)}}}

The endpoints of the transverse axes are the two vertices.
From the graph, we see they are (0,3), and (8,3).

The endpoints of the conjugate axes are the two co-vertices.
From the graph, we see they are (4,-2), and (4,8).

Next, we draw the defining rectangle around the transverse
and conjugate axes like this.  And we draw and extend the
two diagonals of the defining rectangle:

{{{drawing(320,420,-4,12,-9,15, 

graph(320,420,-4,12,-9,15),

blue(line(0,3,8,3), line(4,-2,4,8)), 

green(line(-5,-8.25,13,14.25),line(-5,14.25,13,-8.25)),

green(line(0,-2,8,-2), line(8,-2,8,8), line(8,8,0,8), line(0,8,0,-2)))}}}

Now we can sketch in the hyperbola approching the two ansymptotes:

{{{drawing(320,420,-4,12,-9,15, 

graph(320,420,-4,12,-9,15),

graph(320,420,-4,12,-9,15,3+(5/4)sqrt(x^2-8x)),

graph(320,420,-4,12,-9,15,3-(5/4)sqrt(x^2-8x)),

blue(line(0,3,8,3), line(4,-2,4,8)), 

green(line(-5,-8.25,13,14.25),line(-5,14.25,13,-8.25)),

green(line(0,-2,8,-2), line(8,-2,8,8), line(8,8,0,8), line(0,8,0,-2)))}}}

Edwin</pre>