Question 732377: for the equation of a hyperbola, find the standard equation, vertices, foci, asymptotes, and graph
1) (x^2/36)- (y^2/49)=1
2) 25x^2-16y^2-400=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! for the equation of a hyperbola, find the standard equation, vertices, foci, asymptotes, and graph
1) (x^2/36)- (y^2/49)=1
2) 25x^2-16y^2-400=0
***
Standard form of equation for a hyperbola with horizontal transverse axis:
 , (h,k)=(x,y) coordinates of center
..
1) (x^2/36)- (y^2/49)=1
center:(0,0)
a^2=36
a=√36=6
vertices: (0±a,0)=(0±6,0)=(6,0) and (-6,0)
..
b^2=49
b=√49=7
..
c^2=a^2+b^2=36+49=85
c=√85≈9.2
foci: (0±c,0)=(0±9.2,0)=(9.2,0) and (-9.2,0)
..
slopes of asymptotes=±b/a=±7/6
Equation of asymptote with slope,m=7/6
y=7x/6
..
Equation of asymptote with slope,m=-7/6
y=-7x/6
..
see graph below:
y=±(49x^2/36-49)^.5
I will let you do the 2nd problem.
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