QJEEAdv16Maadbd: JEE Advanced 2016 Mathematics - Class 11 JEEAdv16Maadbd Solution: Let RS be the diameter of the circle , where S is the point (1, …

Let, P $\equiv$ (cos $\theta$ , sin $\theta$ )

$\therefore$ equation of tangent and normal at P

x cos $\theta$ + y sin $\theta$ = 1 ..... (1)

and y = x tan $\theta$ ...... (2)

Now, equation of tangent at S : x = 1 ...... (3)

Solving (1) and (3), Q $\equiv$ (1, cosec $\theta$ $-$ cot $\theta$ )

$\therefore$ equation of straight line parallel to RS drawn from Q

y = cosec $\theta$ $-$ cot $\theta$ ..... (4)

Let, E $\equiv$ (h, k)

$\therefore$ k = h tan $\theta$ [from (2)]

or, $\tan \theta = {k \over h}$

Again, k = cosec $\theta$ $-$ cot $\theta$ [from (4)]

or, $k = {{1 - \cos \theta } \over {\sin \theta }}$

or, $k = {{1 - {h \over {\sqrt {{h^2} + {k^2}} }}} \over {{k \over {\sqrt {{h^2} + {k^2}} }}}} = {{\sqrt {{h^2} + {k^2}} - h} \over k}$

or, ${k^2} = \sqrt {{h^2} + {k^2}} - h$

or, $h + {k^2} = \sqrt {{h^2} + {k^2}}$

$\therefore$ locus of E $x + {y^2} = \sqrt {{x^2} + {y^2}}$

Clearly, points $\left( {{1 \over 3},{1 \over {\sqrt 3 }}} \right)$ and $\left( {{1 \over 3}, - {1 \over {\sqrt 3 }}} \right)$ are on locus of E.

Therefore, (A) and (C) are the correct options.