Show that the curves `(x^2)/(a^2+lambda_1)=1` and `(x^2)/(a^2+lambda_2)+(y^2)/(b^2+lambda_2)=1` intersect at right angles.

Show that the curves (x^2)/(a^2+lambda_1)+(y^2)/(b^2+lambda_1)=1 and (x^2)/(a^2+lambda_2)+(y^2)/(b^2+lambda_2)=1 intersect at right angles.

Tangents,one to each of the ellipses (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and (x^(2))/(a^(2)+lambda)+(y^(2))/(b^(2)+lambda)=1 are drawn Tangents,If the tangents meet at right angles then the locus of their point of intersection is

The orthogonal trajectories of the curves (x^(2))/(a^(2)+lambda)+(y^(2))/(b^(2)+lambda)=1 , lambda being the parameter of the family is

The locus of the point of intersection of perpendicular tangents to x^(2)/a^(2) + y^(2)/b^(2) = 1 and (x^(2))/(a^(2) + lambda) + (y^(2))/(b^(2) + lambda) = 1 , is

The locus of the point of intersection of perpendicular tangents to x^(2)/a^(2) + y^(2)/b^(2) = 1 and (x^(2))/(a^(2) + lambda) + (y^(2))/(b^(2) + lambda) = 1 , is

Prove that the family of curves x^2/(a^2+lambda)+y^2/(b^2+lambda)=1 , where lambda is a parameter, is self orthogonal.

If the lines (x-1)/(1)=(y-3)/(1)=(z-2)/(lambda) and (x-1)/(lambda)=(y-3)/(2)=(z-4)/(1) intersect at a point, then the value of lambda^(2)+4 is equal to

If the lines (x-1)/(1)=(y-3)/(1)=(z-2)/(lambda) and (x-1)/(lambda)=(y-3)/(2)=(z-4)/(1) intersect at a point, then the value of lambda^(2)+4 is equal to

The differential equation satisfying the curve (x^(2))/(a^(2)+lambda)+(y^(2))/(b^(2)+lambda)=1 where lambda be arbitary uknown, is (a) (x+yy_(1))(xy_(1)-y)=(a^(2)-b^(2))y_(1) (b) (x+yy_(1))(xy_(1)-y)=y_(1) ( c ) (x-yy_(1))(xy_(1)+y)(a^(2)-b^(2))y_(1) (d) None of these

The differential equation satisfying the curve (x^(2))/(a^(2)+lambda)+(y^(2))/(b^(2)+lambda)=1 when lambda begin arbitary uknowm, is (a) (x+yy_(1))(xy_(1)-y)=(a^(2)-b^(2))y_(1) (b) (x+yy_(1))(xy_(1)-y)=y_(1) ( c ) (x-yy_(1))(xy_(1)+y)(a^(2)-b^(2))y_(1) (d) None of these