Question 1071603
<pre>
{{{drawing(320,400,-2,10,-8,7,locate(4,-1,"(4,-1)"),
circle(4,-1,.1),green(line(4,4,4,-6)),

 

blue(line(7/4,4,25/4,4),line(7/4,-6,25/4,-6)),

 

locate(.2,3.9,"(7/4,4)"),
circle(4,4,.1), locate(4,4,"(4,4)"),
circle(4,-6,.1), locate(4,-6,"(4,-6)"),
graph(320,400,-2,10,-8,7,(-4/3)sqrt(x^2-8x+25)-1),
graph(320,400,-2,10,-8,7,(4/3)sqrt(x^2-8x+25)-1)   )}}}

The green line above, drawn between foci, is given to 
be 10 units long.  The center is its midpoint, so the 
two foci are (4,4) and (4,-6).
The two blue lines are the latus rectums. They are given
as 9/2, so by subtraction of half that or 9/4 from
the x-coordinate of the focus (4,4), we get that
the left end of the upper latus rectum is the point
(7/4,4).  The hyperbola goes through that point.

We know that the equation of the hyperbola is of the
form

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

and since the center is (4,-1), it's

{{{(y+1)^2/a^2-(x-4)^2/b^2}}}{{{""=""}}}{{{1}}}
or

{{{b^2(y+1)^2-a^2(x-4)^2}}}{{{""=""}}}{{{a^2b^2}}}

Since it goes through (7/4,4), we substitute that 
for (x,y) 

{{{b^2(4+1)^2-a^2(7/4-4)^2}}}{{{""=""}}}{{{a^2b^2}}}

{{{b^2(5)^2-a^2(-9/4)^2}}}{{{""=""}}}{{{a^2b^2}}}

{{{25b^2-a^2(81/16)}}}{{{""=""}}}{{{a^2b^2}}}

{{{400b^2-81a^2}}}{{{""=""}}}{{{16a^2b^2}}}

We know that c = 5 because c is the distance from the
center to the focus.

For all hyperbolas, {{{c^2}}}{{{""=""}}}{{{a^2+b^2}}} or {{{5^2}}}{{{""=""}}}{{{a^2+b^2}}},
and so {{{a^2}}}{{{""=""}}}{{{25-b^2}}}

Substitute in

{{{400b^2-81a^2}}}{{{""=""}}}{{{16a^2b^2}}}

{{{400b^2-81(25-b^2)}}}{{{""=""}}}{{{16(25-b^2)b^2}}}

{{{400b^2-2025+81b^2}}}{{{""=""}}}{{{16b^2(25-b^2)}}}

{{{400b^2-2025+81b^2}}}{{{""=""}}}{{{400b^2-16b^4)}}}

{{{481b^2-2025}}}{{{""=""}}}{{{400b^2-16b^4)}}}

{{{81b^2-2025}}}{{{""=""}}}{{{-16b^4)}}}

{{{16b^4+81b^2-2025}}}{{{""=""}}}{{{0)}}}

{{{(b^2-9)(16b^2+225)}}}{{{""=""}}}{{{0}}}

{{{b^2-9}}}{{{""=""}}}{{{0}}}; {{{16b^2+225}}}{{{""=""}}}{{{0}}}

{{{b^2}}}{{{""=""}}}{{{9}}};  {{{16b^2}}}{{{""=""}}}{{{-225}}}

{{{b^2}}}{{{""=""}}}{{{"" +- 9}}};  {{{b^2}}}{{{""=""}}}{{{-225/16}}}

bē can only be positive, so

bē = 9 

Substitute in 

{{{a^2}}}{{{""=""}}}{{{25-b^2}}}

{{{a^2}}}{{{""=""}}}{{{25-9}}}

{{{a^2}}}{{{""=""}}}{{{16}}}

So the equation:

{{{(y+1)^2/a^2-(x-4)^2/b^2}}}{{{""=""}}}{{{1}}}

becomes:

{{{(y+1)^2/16-(x-4)^2/9}}}{{{""=""}}}{{{1}}}
 
Edwin</pre>