# SOLUTION: 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

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: 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      Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Algebra: Conic sections - ellipse, parabola, hyperbola Solvers Lessons Answers archive Quiz In Depth

 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 =1Answer by lwsshak3(6469)   (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