Find the coordinates of the center, the foci, and the vertices, sketch the curve, and determine the eccentricity of the ellipse given by the equation
![\[ \frac{x^2}{100} + \frac{y^2}{36} = 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ead6a20f5ed9318f9ee9edf6f1b7afd0_l3.png "Rendered by QuickLaTeX.com")
This is an equation for an ellipse in standard form with center at 
.
![\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-56303fecfb5cc71fd865bc5bba607820_l3.png "Rendered by QuickLaTeX.com")
with 
and 
. We then know its foci are located 
and 
where
![\[ c = \sqrt{a^2 - b^2} = \sqrt{100 - 36} = \sqrt{64} = 8. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-74d526f20213ebd195f3f95b72c959a6_l3.png "Rendered by QuickLaTeX.com")
Therefore, the foci of this ellipse are 
and 
. Further, the vertices are given by 
(page 505 of Apostol) so they are at the points 
and 
. Finally, we can compute the eccentricity
![\[ c = |a| e \quad \implies \quad 8 = 10e \quad \implies \quad e = \frac{4}{5}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-590903b3723d666bf1047c8d34324f3c_l3.png "Rendered by QuickLaTeX.com")
from the book the vertices should be at x=a/e and x=-a/e , in this case it should be at x=10*5/4=10.5 and x=-10*5/4=-12.5
im sorry i confused it with vertical lines equation
This makes sense to me, but the vertirx doesnt seem to be equidistant to the line and focus (ie (-10,0) with x = -12.5 and (-8,0). Could someone explain what’s happening here please?
Nevermind, I got it. The distances shouldn’t be equal, but should have a constant ratio represented by the eccentricity (in this case 4/5, or 2/2.5 in what I was observing)