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Question

If x+y2=22 is a tangent to the ellipse x2+2y2=4, then the eccentric angle of the point of contact is
  1. π2
  2. π3
  3. π6
  4. π4


Solution

The correct option is D π4
Any point P(θ) on the ellipse x24+y22=1can be written as (2cosθ,2sinθ).

Tangent at P:
xcosθ2+ysinθ2=1x(2cosθ)+y(2sinθ)=22(ii)

Also, tangent at P is given as
x+y2=22(ii)

If line (i) and (ii) are identical, then
2cosθ1=2sinθ2tanθ=1
θ=π4 (as P lies in first quadrant)

Alternate Solution:
Given line can be rewritten as 
2x4+y2=1
and ellipse is x24+y22=1
Now using the point form of tangent, the point of contact will be (2,1)
Now from the parametric coordinates
acosθ=2, bsinθ=1cosθ=12, sinθ=12θ=π4

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