How do you graph the ellipse #x^2/169+y^2/25=1# and find the center, the major and minor axis, vertices, foci and eccentricity?

1 Answer
Aug 6, 2017

Answer:

Please see below.

Explanation:

This is the equation of an ellipse of the form #(x-h)^2/a^2+(y-k)^2/b^2=1#, whose center is #(h,k)#, major axis is #2a# and minor axis is #2b#. Vertices on major axis are #(h+-a,k)# and along minor axis are #(h,k+-b)#. Eccentricity is given by #e=sqrt(1-b^2/a^2)# and focii are #(h+-ae,k)#.

As #x^2/169+y^2/25=1# can be written as

#(x-0)^2/13^2+(y-0)^2/5^2=1#

Hence, this is an equation of an ellipse, whose center is #(0,0)#, major axis is #13xx2=26# and minor axis is #5xx2=10#.

Vertices are #(-13,0),(13,0),(5,0)# and #(-5,0)#.

Eccentricity is #e=sqrt(1-5^2/13^2)=sqrt(12^2/13^2)=12/13#

and fociie are #(+-(13xx12/13),0)# i.e. #(-12,0)# and #(12,0)#.

We can mark the four vertices, if so desired a few more points by using the equation #x^2/169+y^2/25=1#. For example, if #x=+-3#. #y=+-5sqrt(1-9/169)=+-20sqrt(10/169)=+-4.865# gives four more points #(3,4.865),(-3,4.865), (3,-4.865)# and #(-3,-4.865)#, joining which along with vertices will give us the desired ellipse.

The ellipse appears as shown below:

graph{(x^2/169+y^2/25-1)((x+13)^2+y^2-0.04)((x-13)^2+y^2-0.04)(x^2+(y+5)^2-0.04)(x^2+(y-5)^2-0.04)((x+12)^2+y^2-0.04)((x-12)^2+y^2-0.04)=0 [-20, 20, -10, 10]}