for x^2 /169 + y^2 /144+1
a. length of semi major axis
b. " " " minor axis
c. coordinates of the foci
d. eccentricity

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1 Answer


a) a=√169=13

b) b=√144=12 (+1 should be =1)

c) focal distance from centre (0,0) is √(169-144)=√25=5. Foci at (-5,0) and (5,0).

d) √(1-144/169)=√(25/169)=5/13.
by Top Rated User (788k points)

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