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Locus of the mirror point P for tangent rays from (cos t, sin t) to (-sin t, cos t) - JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
A ray starting from the point A(cos⁡θ,sin⁡θ)A(\cos\theta,\sin\theta) strikes a mirror at a variable point PP and, after reflection, passes through the point B(−sin⁡θ,cos⁡θ)B(-\sin\theta,\cos\theta), the incident ray and the reflected ray both being tangents to the circle x2+y2=1x^2+y^2=1. The variable point PP then traces x2+y2=kx^2+y^2=k. Find the value of kk.
Solution
Answer: 2
Question attachment Step 1:
cos⁡2θ+sin⁡2θ=1,(−sin⁡θ)2+cos⁡2θ=1\cos^2\theta+\sin^2\theta = 1 ,\qquad (-\sin\theta)^2+\cos^2\theta = 1
 so AA and BB both lie on x2+y2=1x^2+y^2=1, and the two rays touch the circle there. Step 2:
OA⃗⋅OB⃗=cos⁡θ(−sin⁡θ)+sin⁡θcos⁡θ=0\vec{OA}\cdot\vec{OB} = \cos\theta(-\sin\theta) + \sin\theta\cos\theta = 0
 so OA⊥OBOA \perp OB for every θ\theta. In particular A≠BA \ne B, and the tangents at AA and BB, being perpendicular to OAOA and OBOB, are not parallel, so they do meet at a point PP. Step 3: A tangent is perpendicular to the radius at its point of contact, so PA⊥OAPA \perp OA and PB⊥OBPB \perp OB. In the quadrilateral OAPBOAPB the angles at OO, AA and BB are 90∘90^\circ each, so
∠APB=360∘−90∘−90∘−90∘=90∘\angle APB = 360^\circ-90^\circ-90^\circ-90^\circ = 90^\circ
 The incident and reflected rays are therefore perpendicular. Step 4: The locus of points from which the two tangents to a circle are perpendicular is that circle's director circle, of radius 2 r\sqrt2\,r. Here r=1r=1, so PP lies on
x2+y2=2x^2+y^2 = 2
 Step 5:
k=2k = 2
 Answer: 22
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