# Find the equation of the ellipse referred to its centre (0,0) (a) whose latus rectum is 5 and whose eccentricity is 2/3. (b) whose minor axis is equal to the distance between the foci and whose latus rectum is 10

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2015-09-16T04:52:07+05:30

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X² / a²  + y² / b²  = 1

where a = semi major axis,    b = semi minor axis.
center of ellipse is  (0,0)  and the ellipse is symmetric about X and Y axes.

Latus Rectum = 2 b² / a = 5
eccentricity :    e = 2/3
so     b² = a² (1 - e²) = a² * 5/9

hence,  a = 2 b² / 5  = 2 a²/9        =>  a = 9/2
b = 3√5 / 2

so equation is :  4 x²/9 +  4 y² /45 = 1
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2)
distance between focii =  2 a e = 2 * b      given
so b = a e

we know that          b² =  a² - a² e²  =  a² - b²
=>  b = a/√2
e = 1/√2

2 b² / a = latus rectum = 10
so  2 * a / 2   =  10
=>  a = 5        and  b  = 5/√2

so  x²/25 + 2 y² /25  = 1

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