If the normals at `P(theta)` and `Q(pi/2+theta)` to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` meet the major axis at `Ga n dg,` respectively, then `P G^2+Qg^2=` `b^2(1-e^2)(2-e)^2` `a^2(e^4-e^2+2)` `a^2(1+e^2)(2+e^2)` `b^2(1+e^2)(2+e^2)`

If the normals at P(theta) and Q((pi)/(2)+theta) to the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 meet the major axis at G and g, respectively,then PG^(2)+Qg^(2)=b^(2)(1-e^(2))(2-e)^(2)a^(2)(e^(4)-e^(2)+2)a^(2)(1+e^(2))(2+e^(2))b^(2)(1+e^(2))(2+e^(2))

If e be the eccentricity of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2)) = 1 , then e =

If the normal at 'theta' on the hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 meets the transverse axis at G , and A and A' are the vertices of the hyperbola , then AC.A'G= (a) a^2(e^4 sec^2 theta-1) (b) a^2(e^4 tan^2 theta-1) (c) b^2(e^4 sec^2 theta-1) (d) b^2(e^4 sec^2 theta+1)

Find the area bounded by the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 and the ordinates x = 0 and x = a e , where, b^2=a^2(1-e^2) and e < 1 .

If the normal at the end of latus rectum of an ellipse x^(2)//a^(2)+y^(2)//b^(2)=1 of eccentricity e passes through one end of the minor axis then e^(4)+e^(2)=

If the normal at the end of latus rectum of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 passes through (0,b) then e^(4)+e^(2) (where e is eccentricity) equals

The normal at an end of a latus rectum of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 passes through an end of the minor axis if (A)e^(4)+e^(2)=1(B)e^(3)+e^(2)=1(C)e^(2)+e=1(D)e^(3)+e=1

If e_(1)ande_(2) be the eccentricities of the ellipses (x^(2))/(a^(2))+(4y^(2))/(b^(2))=1and(x^(2))/(a^(2))+(4y^(2))/(b^(2))=1 respectively then prove that 3=4e_(2)^(2)-e_(1)^(2) .

If e_(1)ande_(2) be the eccentricities of the ellipses (x^(2))/(a^(2))+(y^(2))/(b^(2))=1and(x^(2))/(a^(2))+(4y^(2))/(b^(2))=1 respectively then prove that 3=4e_(2)^(2)-e_(1)^(2) .

If omega is one of the angles between the normals to the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 at the point whose eccentric angles are theta and pi/2+theta , then prove that (2cotomega)/(sin2theta)=(e^2)/(sqrt(1-e^2))