Question

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

Answer 1

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