Dear student, As E:x2a2+y2b2=1, if b→a, the e becomes 0 as e=a2-b2a2 as b→a, e→0so the focii (±ae,0), becomes (0,0) and the Directrix that is x=±ae which receded to ∞when e→0so option A is correct Hope this clears your doubt With regards

Plzzz solve this one

Dear student,

A s E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, i f b \to a, t h e e b e c o m e s 0 a s e = \sqrt{\frac{a^{2} - b^{2}}{a^{2}}} a s b \to a, e \to 0 s o t h e f o c i i (\pm a e, 0), b e c o m e s (0, 0) a n d t h e D i r e c t r i x t h a t i s x = \pm \frac{a}{e} w h i c h r e c e d e d t o \infty w h e n e \to 0 s o o p t i o n A i s c o r r e c t .

Hope this clears your doubt
With regards

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