How do you find the vertex, the directrix, the focus and eccentricity of: #(9x^2)/25 + (4y^2)/25 = 1#?

Question

2884 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-16T05:33:00

Please see below.

This a typical equation of an ellipse centered at origin as it is of the form #x^2/a^2+y^2/b^2=1#, as we can write it as

#x^2/(5/3)^2+y^2/(5/2)^2=1#

As #b>a#, we have major axis #5/2xx2=5# along #y# axis and minor axis #5/3xx2=10/3# along #x# axis.

Vertices, along major axis, are #(0,+-2.5)# and co-vertices, along minor axis, are #(+-1.67,0)#

and eccentricity given by #a^2=b^2(1-e^2)# i.e. #e=sqrt(1-a^2/b^2)# is

#e=sqrt(1-25/9xx4/25)=sqrt(1-4/9)=sqrt5/3=0.7454#

Focii, along #y#-axis, are given by #(0,+-be)# i.e. #(0,+-1.86)#

and the two directrix are #y=+-b/e=+-3.354#

graph{(9x^2+4y^2-25)(x^2+(y-1.86)^2-0.005)(x^2+(y+1.86)^2-0.005)=0 [-5.02, 4.98, -2.36, 2.64]}