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

Options

A. `y^(2)=2x`
B. `y^(2)=1-2x`
C. `2x=y^(2)-1`
D. `y^(2)=1-x`

Answer

B

Explanation

Let `P(cos theta, sin theta)` be a point on the circle `x^(2)+y^(2)=1`. Then, the equations of the tangent and the normal to the circle at `P(cos theta, sin theta)` are

`x cos theta + y sin theta = 1 ...(i)`
and, `x sin theta - y cos theta = 0 ...(ii)`
The tangent to the circle at S(1, 0) is x=1. ...(iii)
Lines (i) and (iii) intersect at Q `(1, (1-cos theta)/(sin theta))`.
The equation of the line through Q parallel to RS is
`y=(1-cos theta)/(sin theta)`. or, `y=tan (theta)/(2)`. This line intersects the normal at P at
point `E((tan theta//2)/(tan theta), tan (theta)/(2))`. Let the coordinates of E be (h, k).
Then,
`h=(tan theta//2)/(tan theta) and k = tan. (theta)/(2)`
`rArr 2h=1 - tan^(2) (theta)/(2) and k = tan. (theta)/(2) rArr 2h = 1 - k^(2)`
Hence, the locus of E (h, k) is `2x=1-y^(2) or, y^(2)=1 - 2x`.