Question

# If x+y√2=2√2 is a tangent to the ellipse x2+2y2=4, then the eccentric angle of the point of contact isπ2π3π6π4

Solution

## The correct option is D π4Any point P(θ) on the ellipse x24+y22=1can be written as (2cosθ,√2sinθ). Tangent at P: xcosθ2+ysinθ√2=1⇒x(√2cosθ)+y(2sinθ)=2√2…(ii) Also, tangent at P is given as x+y√2=2√2…(ii) If line (i) and (ii) are identical, then √2cosθ1=2sinθ√2⇒tanθ=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=1⇒cosθ=1√2, sinθ=1√2⇒θ=π4

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