Question 627514
Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the equation y²/25 - x²/16 . Then graph the hyperbola.
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{{{y^2/25-x^2/16=1}}}
This is an equation of a hyperbola with vertical transverse axis.
Its standard form: {{{(y-k)^2/a^2-(x-h)^2/b^2=1}}}, (h,k)=(x,y) coordinates of center.
For given equation:
center: (0,0)
a^2=25
a=√25=5
Vertices: (0,0±a)+(0,0±5)=(0,-5) and (0,5)
b^2=16
b=√16=4
c^2=a^2+b^2=25+16=41
c=√41≈6.4
Foci:  (0,0±c)+(0,0±6.4)=(0,-6.4) and (0,6.4)
..
slopes of asymptotes=±a/b=±5/4
Equations of asymptotes:
y=-5x/4
and 
y=5x/4
..
See graph below
y=±(25+25x^2/16)^.5

{{{ graph( 300, 300, -10,10, -10, 10,(25+25x^2/16)^.5,-(25+25x^2/16)^.5,-5x/4,5x/4) }}}