An ellipse of eccentricity 
is inscribed in an ellipse of equal eccentricity and area equals to 9 square units in such a way that both the ellipse touch each other at end of their common major axis. If length of major axis of smaller ellipse is equal to length of minor axis of bigger ellipse, find the area of the bigger ellipse outside the smaller ellipse.
Text Solution
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Ans. 0004
Sol. Let the ellipse be 
+ 
= 1
and 
+ 
= 1
Required area = π ab – π bc
= π b (a – c)
b 2 = a 2 (1– e 2 ) and c 2 = b 2 (1 – e 2 )
⇒ c 2 = a 2 (1 – e 2 ) 2 ⇒ c = a (1 – e
2 ) ⇒ a – c = ae 2 so required area = π b (ae
2 ) = π abe 2 = 9 × 
= 4 sq. units.
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