Question

# A coplanar beam of light emerging from a point source have the equation λx–y+2(1+λ)=0,∀ λ∈R; the rays of the beam strike an elliptical surface and get reflected inside the ellipse. The reflected rays form another convergent beam having the equation μx−y+2(1−μ)=0,∀ μ∈R. Further it is found that the foot of the perpendicular from the point (2,2) upon any tangent to the ellipse lies on the circle x2+y2–4y–5=0.The eccentricity of the ellipse is equal to 23.The eccentricity of the ellipse is equal to 13 The area of the largest triangle that an incident ray and corresponding reflected ray can enclose with major axis of the ellipse is equal to 2√5.The area of the largest triangle that an incident ray and corresponding reflected ray can enclose with major axis of the ellipse is equal to 4√5

Solution

## The correct options are A The eccentricity of the ellipse is equal to 23. C The area of the largest triangle that an incident ray and corresponding reflected ray can enclose with major axis of the ellipse is equal to 2√5.(2−y)+λ(x+2)=0⇒ family of lines through (−2,2) And (2,−y)+μ(x−2)=0⇒ family of lines tharough (2,2) and x2+y2−4y−5=0 will be auxilary circle & its radius =a=3 Distance between foci =2ac=4 ⇒ae=2 ∴e=23 Area of ΔPF1F2=12×base×height=12×4×height maximum area = maximum height = b ⇒b2=a2(1−e2)=9(1−49) ⇒b=√5 Area=2√5

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