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MathematicsMultiple correct+4 / -2Question 49 of 54

JEE Advanced 2016 · Paper 1, Mathematics question 49

Let RSRS be the diameter of the circle x2+y2=1x^2+y^2=1, where SS is the point (1,0)(1,0). Let PP be a variable point (other than RR and SS) on the circle and tangents to the circle at SS and PP meet at the point QQ. The normal to the circle at PP intersects a line drawn through QQ parallel to RSRS at point EE. Then the locus of EE passes through the point(s)

Choose one or more options
Answer and worked solution

Official answer: A, C

Set up. Let P=(cos⁡θ,sin⁡θ)P=(\cos\theta,\sin\theta). The tangent at SS is x=1x=1, and the tangent at PP is xcos⁡θ+ysin⁡θ=1x\cos\theta+y\sin\theta=1. They meet at
Q=(1,tan⁡θ2).Q=\left(1,\tan\frac\theta2\right).

Point E. The line through QQ parallel to RSRS is y=ty=t, with t=tan⁡θ2t=\tan\frac\theta2. The normal at PP is y=xtan⁡θy=x\tan\theta. So
E=(ttan⁡θ,t)=(1−t22,t).E=\left(\frac{t}{\tan\theta},t\right)=\left(\frac{1-t^2}2,t\right).

Locus. y2=1−2xy^2=1-2x.

  • (13,±13)\left(\frac13,\pm\frac1{\sqrt3}\right): 13=1−23\frac13=1-\frac23. ✓
  • (14,±12)\left(\frac14,\pm\frac12\right): 14≠1−12\frac14\ne1-\frac12. ✗

(A), (C)

Topics:LocusTangents and normals
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