for x^2 /169 + y^2 /144=1
find
a. length of semi major axis
b. " " " minor axis
c. coordinates of the foci
d. eccentricity
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x^2/169+y^2/144=1 is the equation of an ellipse, which can be written: x^2/13^2+y^2/12^2=1. The semi-major axis is 13 (horizontal "radius") and the semi-minor axis is 12 (vertical "radius").

The origin of the ellipse is (0,0). The foci are found by invoking the properties of an ellipse which is that for all points on the ellipse the sum of the distances from each point to each focus is constant. The distance of each focus from the origin is the same, one sitting on the x axis on the left of the origin, the other on the right, a distance f. So the coordinates are (-f,0) and (f,0). To find f we note first that the x intercepts are 13 and -13. Take (13,0), for example, the distance between the right focus and the intercept is 13-f. The distance between the same intercept and the other focus is 13+f. The sum of these two distances is 26, the length of the major axis. Now take the y intercept (0,12). This point is sqrt(12^2+f^2) from the right focus and the same distance from the left focus. The sum of these distances must be 26, so 2sqrt(12^2+f^2)=26 and 12^2+f^2=13^2, f^2=169-144=25, and f=5. So the coordinates of the foci are (5,0) and (-5,0). 

The eccentricity is the ratio of the distance between the foci (10) to the length of the major axis (26)=10/26=5/13.

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