Let RS be the diameter of the circle `x^2+y^2=1,` where S is the point `(1,0)` Let P be a variable apoint (other than `R and S`) on the circle and tangents to the circle at `S and P` meet at the point Q.The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. then the locus of E passes through the point(s)-

Options

A. `((1)/(3),(1)/(sqrt3))`
B. `((1)/(4),(1)/(2))`
C. `((1)/(3),-(1)/(sqrt3))`
D. `((1)/(4),-(1)/(2))`

Answer

A, C

Explanation

Given, RS is the diameter of `x^(2)+y^(2)=1`.
Here, equation of the tangent at `P(costheta,sintheta) " is "xcostheta+ysintheta=1`.

Intersecting with x = 1,
`y=(1-costheta)/(sintheta)`
`therefore Q(1(1-costheta)/(sintheta))`
`therefore` Equation of the line through Q parallel to RS is
`y=(1-costheta)/(sintheta)= (2 sin^(2) ""(theta)/(2))/ (2 sin""(theta)/(2)cos""(theta)/(2)) = tan""(theta)/(2) " "...(i)`
Normal at `P:y=(sintheta)/(costheta)*x`
`rArr y = x tantheta " "...(ii)`
Let their point of intersection be (h, k).
Then, `h=tan""(theta)/(2)and k=htantheta`
`therefore k=h((2tan""(theta)/(2))/(1-tan^(2)""(theta)/(2)))rArr k=(2h-k)/(1-k^(2))`
`rArrk(1-k^(2))=2hk`
`therefore " Locus for point E ":2x=(1-y^(2))" "...(iii)`
When `x=(1)/(3)`, then
`1-y^(2)=(2)/(3)rArr y^(2)=1-(2)/(3)rArr y=pm(1)/(sqrt3)`
`therefore ((1)/(3),pm(1)/(sqrt3)) " satisfy " 2x=1-y^(2)`.
when `x=(1)/(4)`, then
`1-y^(2)=(2)/(4)rArr y^(2)=1-(1)/(2)rArr y=pm(1)/(sqrt2)`
`therefore ((1)/(4),pm(1)/(2)) " does not satisfy " 1-y^(2)=2x`.