" 1"(103.49)/(6)times sin theta-y sin theta=sqrt(x^(2)-y^(2))" ait "(cos^(2)theta)/(a^(2))+(sin^(2)theta)/(b^(2))=(1)/(x^(2)+y^(2))

Eliminate theta between x sin theta-ycos theta =sqrt(x^2+y^2) and (cos^2theta)/(a^2)+(sin^2theta)/(b^2)=1/(x^2+y^2) .

If x sin theta -y cos theta =sqrt(x^2 +y^2) and cos^2 theta/a^2 +sin^2 theta/b^2 =1/(x^2 +y^2) , then the correct relation is

x=b sin^(2)theta & y=a cos^(2)theta

A hyperbola having the transverse axis of length 2sin theta is confocal with the ellipse 3x^(2)+4y^(2)=12 .Then its equation is x^(2)csc^(2)theta-y^(2)sec^(2)theta=1x^(2)sec^(2)theta-y^(2)cos ec^(2)theta=1x^(2)sin^(2)theta-y^(2)cos^(2)theta=1x^(2)cos^(2)theta-y^(2)sin^(2)theta=1

If cos^(-1)x+cos^(-1)y=theta show that x^(2)-2xy cos theta+y^(2)=sin^(2)theta

If x= a cos^(2) theta sin theta and y= a sin^(2) theta cos theta , " then " ((x^(2)+y^(2))^(3))/(x^(2)y^(2))=

If x=a sin theta and y=b cos theta , then prove : (x^(2))/(a^(2))+(y^(2))/(b^(2))=1

If x=a (theta-sin theta) and y=a(1-cos theta) , find (d^(2)y)/(dx^(2)) at theta=pi.

Angle between the lines x^(2)[cos^(2) theta-1]-xy sin^(2) theta+y^(2) sin^(2) theta= theta is