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Mathematics 14 Online
darkknight:

how would you write the asymptote of this function (y^2)/49)-(x^2)/121)=1

jabari2020:

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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