The normal to the ellipse `4x^(2)+5y^(2)=20` at the point `P` touches the parabola `y^(2)=4x` .The eccentric angle of the point `P` is:

The tangent and the normal to the ellipse 4x^(2)+9y^(2)=36 at the point P meets the major axis Q and R respectively.If QR=4 then the eccentric angle of P is given by

If x+y=k is a normal to the parabola y^(2)=12x, then it touches the parabola y^(2)=px then |P| =

The given circle x^(2)+y^(2)+2px=0 , p in R touches the parabola y^(2)=-4x externally then p =

If the normal to the ellipse 3x^(2)+4y^(2)=12 at a point P on it is parallel to the line , 2x+y=4 and the tangent to the ellipse at P passes through Q (4,4) then Pq is equal to

The distance of a point on the ellipse (x^(2))/(6)+(y^(2))/(2)=1 from the centre is 2 Then the eccentric angle of the point is

The equation of normal to the ellipse x^(2)+4y^(2)=9 at the point wherr ithe eccentric angle is pi//4 is

Tangents are drawn to the ellipse (x^(2))/(36)+(y^(2))/(9)=1 from any point on the parabola y^(2)=4x. The corresponding chord of contact will touch a parabola,whose equation is

If the distance of a point on the ellipse (x^(2))/(6) + (y^(2))/(2) = 1 from the centre is 2, then the eccentric angle of the point, is