Radius =3a Length of major axis =4c Now, (Radius )< (Half of the length of major axis) 3a<2c9a2<4c29ac−9a2>9ac−4c2
9ac−9a2−2c2>9ac−6c2 Again 3a<2c ⇒9ac<6c2 ⇒9ac−6c2<0 From (i) and (ii), 9ac−9a2−2c2>0
If a and c are positive real numbers and the ellipse 4c2x2+c2y2=1 has four distinct points ir common with the circle x2+y2=9a2, then
Held on 9 Apr 2013 · Verified 6 Jul 2026.
9ac−9a2−2c2<0
6ac+9a2−2c2<0
9ac−9a2−2c2>0
6ac+9a2−2c2>0
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