SOLUTION: Find the foci for each hyperbola and sketch a graph Y^2/49-x^2/16=1 5x^2-10y^2=80

Question 434883: Find the foci for each hyperbola and sketch a graph
Y^2/49-x^2/16=1
5x^2-10y^2=80
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find the foci for each hyperbola and sketch a graph
Y^2/49-x^2/16=1
5x^2-10y^2=80
..
Standard form of hyperbola with vertical transverse axis: (y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
Standard form of hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) again being the (x,y) coordinates of the center.
..
Y^2/49-x^2/16=1
This is a hyperbola with center at (0,0) and with a vertical transverse axis.
a^2=49
a=7 (distance from center to vertices)
b^2=16
b=4
c^2=a^2+b^2=49+16=65
c=sqrt(65)=8.06..(distance from center to foci)
Equation of asymptotes:y=+-(a/b)x=+-(7/4)x
see the graph of the hyperbola below:
..
y=+-(49((x^2/16)+1))^.5

..
5x^2-10y^2=80
x^2/16-y^2/8=1
This is a hyperbola with center at (0,0) but with a horizontal transverse axis.
a^2=16
a=4 (distance from center to vertices)
b^2=8
b=sqrt(8)=2.83
c^2=a^2+b^2=16+8=24
c=sqrt(24)=4.9(distance from center to foci)
Equation of asymptotes:y=+-(b/a)x=+-(2.83/4)x
see the graph of the hyperbola below:
..
y=+-(x^2/2-8)^.5

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