We are given:
eccentricity e = 1 2 e = \frac{1}{\sqrt{2}} e = 2 1 Length of the latus rectum L = 14 L = \sqrt{14} L = 14 We know the formula for the latus rectum of an ellipse is:
L = 2 b 2 a L = \frac{2b^2}{a} L = a 2 b 2
From the problem, we are given L = 14 L = \sqrt{14} L = 14 , so:
2 b 2 a = 14 \frac{2b^2}{a} = \sqrt{14} a 2 b 2 = 14
Thus,
2 b 2 = a 14 (1) 2b^2 = a \sqrt{14} \tag{1} 2 b 2 = a 14 ( 1 )
Next, use the relationship for the eccentricity e e e of an ellipse:
e = 1 − b 2 a 2 = 1 2 e = \sqrt{1 - \frac{b^2}{a^2}} = \frac{1}{\sqrt{2}} e = 1 − a 2 b 2 = 2 1
Squaring both sides:
e 2 = 1 − b 2 a 2 = 1 2 e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2} e 2 = 1 − a 2 b 2 = 2 1
This implies:
b 2 a 2 = 1 2 \frac{b^2}{a^2} = \frac{1}{2} a 2 b 2 = 2 1
Substitute this into equation (1):
2 ( a 2 2 ) = a 14 2 \left(\frac{a^2}{2}\right) = a \sqrt{14} 2 ( 2 a 2 ) = a 14 a 2 = a 14 a^2 = a \sqrt{14} a 2 = a 14
Thus:
a = 14 a = \sqrt{14} a = 14 1
Substitute a = 14 a = \sqrt{14} a = 14 into b 2 = a 2 2 b^2 = \frac{a^2}{2} b 2 = 2 a 2 :
b 2 = 14 2 = 7 b^2 = \frac{14}{2} = 7 b 2 = 2 14 = 7
Now, compute the square of the eccentricity:
e 2 = 1 − b 2 a 2 = 1 − 7 14 = 7 14 = 1 2 e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{7}{14} = \frac{7}{14} = \frac{1}{2} e 2 = 1 − a 2 b 2 = 1 − 14 7 = 14 7 = 2 1
Thus, the square of the eccentricity is 3 2 \frac{3}{2} 2 3 .