Question 601326: the directions are to graph the equation. identify the vertices, foci, and asymptotes or the hyperbola. the problem is:
y^2/49 - x^2/121 =1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! the directions are to graph the equation. identify the vertices, foci, and asymptotes or the hyperbola. the problem is:
y^2/49 - x^2/121 =1
This is an equation for 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: y^2/49 - x^2/121 =1
center: (0,0)
a^2=49
a=√49=7
vertices: (0,0±a)=(0,0±7)=(0,-7) and (0,7)
..
b^2=121
b=√121=11
..
c^2=a^2+b^2
c^2=49+121
c=√170≈13.04
Foci: (0,0±c)=(0,0±13.04)=(0,-13.04) and (0,13.04)
..
Asymptotes:
Slope of asymptotes: ±a/b=±7/11
Asymptotes are straight lines that go thru the center (0,0)
Standard form of equation: y=mx+b, m=slope, b=y-intercept
y-intercept=0
Equations:
y=7x/11
and
y=-7x/11
see graph below:
y=±(49+49x^2/121)^.5
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