The distance of a point on the ellipse `x^2/6 + y^2/2 = 1` from the centre is 2. Find the eccentric angle of the point.

The distance of point on the ellipse x^2 + 3y^2 = 6 from the centre is 2. Find the eccentric angle of the point in the first quadrant. Also find the equation of the tangent at the point.

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

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

The distance of a point on the ellipse (x^(2))/(6)+(y^(2))/(2)=1 from the center is 2. Then the eccentric angle of the point is (pi)/(4) (b) (3 pi)/(4) (c) (5 pi)/(6) (d) (pi)/(6)

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

The eccentric angle of a point on the ellipse x^2/6 + y^2/2 = 1 whose distance from the centre of the ellipse is 2, is (A) pi/4 (B) 7pi/6 (C) 3pi/2 (D) 5pi/3

The distance of the point 'theta' on the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 from a focus, is

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

Find the eccentric angle of a point on the ellipse x^2 + 3y^2 = 6 at a distance 2 units from the centre of the ellipse.