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ur ans's right after all..its just a little bit of rearrangement..put 2=e^2=(a^2+b^2)/ a^2..u get the ans..

ans's x = y = +/- (a^2 + b^2) / 2a

brilliant..thnks dude..

pls try..n share wtever ans ur gettin..

any clue anyone??

The set of positive values of 'm' fr which a line with slope 'm' is a common tangent to ellipse x^2/b^2 + y^2/c^2 =1 n the parabola y^2=4ax is:

(a) (3,5)

(b) (2,3)

(c) (1,3)

(d) (0,1)

???

there's no root there rhd..but ans's close..wts ur logic??

right ans..'m gettin the 2 coordinates opp to wt u gt..n denominator pt of intersection also as sin(w-v)/2..but ans is straight line..right on..

bulls eye..gr8 answer..

sure thing..right now i stil have few doubts floatin in this forum..y dont u try those..

thnks again..ans is right..opt c..

hey isnt that the condition fr line to be tangent of ellipse?? i think the eqn to normal given as a^2 x/ x1 - b^2 y/ y1 = a^2- b^2 wud help cuz by ur method 'm not gettin a a^2- b^2 term as needed by the options.. here x1 n y1 lie on the ellipse..but how do u eliminate them both is the ques..

thnks a lot..ans is right..opt c..