If `s and S'` are the focii BB' is the minor axis such that `/_SBS_1 = sin^(-1) (3/5)` then e =

If S and S^(1) are the foci BB^(1) is the minor axis such that Ang(SBS^(1)) = sin^(-1) (3/5) then e =

Axes are coordinate axes,S and S' are foci,"B" and "B'" are the ends of minor axis angle SBS'=sin^(-1)((4)/(5)) " .Area of "SBS'B'" is 20 sq.units.,then the equation of the ellipse is

Axes are co-ordinate axes, S and S^(1) are Foci, B and B^(1) are the ends of minor axis |SBS^(1) = sin ^(-1) ((4)/(5) ) . if area of SB S^(1) B^(1) is 20sq. Units then eq. of the ellipse is

Let S, S^(') are the focii and BB^(') be the minor axis of an ellipse. If angleBSS^(')=theta then its eccentricity is

S and T are the foci of an ellipse and B is an end point of the minor axis . IF /_\STB is equilateral then e =

S_(1) and S_(2) are the foci of an ellipse and B is the end of the minor axis. If the triangle S_(1) S_(2) B is an equilateral triangle, then eccentricity of the ellipse is

One of extremity of minor of ellipse is B . S and S\' are focii of ellipse then area of right angle /_\SBS\' is equal to 8. Then latus rectum of ellipse is equal to

Let S, S' be the focii and B, B' be the minor axis of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 if angle BSS'= theta and eccentrictiy of the ellipse is e, then show that e=cos theta

Let S, S' be the focil and BB' be the minor axis of the ellipse (x^(2))/(a^(2)) + (y^(2))/(b^(2)) = 1. If angle BSS' = theta , then the eccentricity e of the ellipse is equal to