SOLUTION: Identify the vertices, foci, and slope of the asymptotes for the hyperbola x^2/25 - y^2/36=1 and Write the following equation in standard form and identify it as a parab

Question 481667: Identify the vertices, foci, and slope of the asymptotes for the hyperbola
x^2/25 - y^2/36=1

and

Write the following equation in standard form and identify it as a parabola, ellipse, circle, or hyperbola: 4x^2+8x-9y^2-36y-68=0

thank you so much
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Identify the vertices, foci, and slope of the asymptotes for the hyperbola
x^2/25-y^2/36=1
and Write the following equation in standard form and identify it as a parabola, ellipse, circle, or hyperbola: 4x^2+8x-9y^2-36y-68=0
***
x^2/25-y^2/36=1
This is an equation of a hyperbola with horizontal transverse axis of the standard form with (h,k) being the center.
(x-h)^2/a^2-(y-k)^2/b^2=1
For given equation:
Center:(0,0)
a^2=25
a=√25=5
length of transverse axis=2a=10
vertices (end-points of transverse axis)=0�a,0)=0�5,0)=(5,0) and (-5,0)
..
b^2=36
b=√36=6
..
c^2=a^2+b^2=25+36=61
c=√61=7.8
Foci: (0�c,0)=0�√61,0)=(√61,0) and (-√61,0)
slope of asymptotes: �b/a=�6/5
...
4x^2+8x-9y^2-36y-68=0
4(x^2+2x+1)-9(y^2+4y+4)=68+4-36=36
4(x+1)^2-9(y+2)^2=36
Divide both sides by 36
(x+1)^/9-(y+2)^2/4=1
This is an equation of a hyperbola with horizontal transverse axis of the standard form:
(x-h)^2/a^2-(y-k)^2/b^2=1
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