The line `x/a cos theta -y/b sin theta=1` will touch the ellipse `x^2/a^2+y^2/b^2=1` at point P whose eccentric angle is (A) `theta` (B) `pi-theta` (C) `pi+theta` (D) `2pi-theta`

if P(theta) and Q(pi/2 +theta) are two points on the ellipse x^2/a^2+y^@/b^2=1 , locus ofmid point of PQ is

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

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

The straight line x cos theta+y sin theta=2 will touch the circle x^(2)+y^(2)-2x=0 if theta=n pi,n in IQ(b)A=(2n+1)pi,n in Itheta=2n pi,n in I(d) none of these

If x = a cos theta - b sin theta and y = a sin theta + b cos theta, prove that x^2 + y^2 = a^2 + b^2.

If x=a cos theta-b sin theta and y=a sin theta+b cos theta then prove that :x^(2)+y^(2)=a^(2)+b^(2)

P and Q are two points on the ellipse x^2/a^2+y^2/b^2=1 with eccentric angles theta_1 and theta_2 . Find the equation of the locus of the point of intersection of the tangents at P and Q if theta_1+theta_2=pi/2 .

If x =a ( theta + sin theta), y =a (1 - cos thea) then (d ^(2) y)/(dx ^(2)) at theta = (pi)/(4) is

Find the slope of the normal to the curve x = 1 - a sin theta, y = b cos^(2) theta at theta = (pi)/2 .

If x=a ( theta - sin theta), y =a (1 - cos theta) then y _(2) at theta = pi //4 is