How do you find the center and vertices of the ellipse #x^2/9+y^2/5=1#?

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Answer 1 · 2018-07-20T13:48:58

Please see the explanation below

Compare this equation to the standard equation of the ellipse

#(x-h)^2/a^2+(y-k)^2/b^2=1#

Our equation is

#x^2/9+y^2/5=1#

#a=sqrt9=+-3#

#b=+-sqrt5#

The center is #C=(h,k)=(0,0)#

The vertices are :

#A=(h,a)=(3,0)#

#A'=(h,-a)=(-3,0)#

#B=(k,b)=(0,sqrt5)#

#B'=(k,-b)=(0,-sqrt5)#

graph{(x^2/9+y^2/5-1)=0 [-7.9, 7.904, -3.95, 3.95]}

Answer 2 · 2018-07-20T13:52:09

See the answers below

The given equation of ellipse:

#x^2/9+y^2/5=1#

#x^2/3^2+y^3/{(sqrt5)^2}=1#

The above equation of ellipse is in form of #x^2/a^2+y^2/b^2=1# which has

Center: #(x=0, y=0)\equiv(0, 0)#

Vertices: #(x=\pm a, y=0)\ \ &\ \ \ (x=0, y=\pmb)#

#(\pm3, 0) \ \ \ &\ \ \ (0, \pm\sqrt5)#