Question

Find the eccentricity of the graph of ((x-2)^2/121) + ((y+6)^2/64)=1 and of the graph of ((x-2)^2/121)-((y+6)^2/64)=1 . Using complete sentences compare the eccentricity of the graph of an ellipse with the graph of a hyperbola.

Answer 1

?? help??

Answer 2

e=c/a
for an ellipse c is the length of the semi major axis and a is the major axis since a is greater than b(the minor axis) and c
a2=b2+c2
for hyperbola
the same formula holds
e=c/a
but this time
c is the largest
thus
c2=a2+b2
to make things easier try to collect variables together and see if it's an equation of an ellipse or hyperbola
The first equation
(x-2)^2/121) + ((y+6)^2/64)=1
is already simplified and
is written in the form
((x-h)^2/a^2)+((y-k)^2/b^2)=1\
now from the given equation
a^2=121
and
b^2=64
thus the ellipse is an x (horizontal ellipse)
to find c use the above formula I gave you
a^2=b^2+c^2
c=sqrt(a^2-b^2)
c=sqrt(121-68)
c=sqrt(53)
and because a^2=121 a=sqrt(121)
e=c/a=(sqrt(53))/(sqrt(121))

use similar steps to find the ecentricity for the hyperbola.but remeber that c(the transverse axis is the largest so instead of the formula a^2=b^2+c^2
you should use c^2=a^2+^2)

Let me know if you have any doubts.