A curve 'C' with negative slope through the point(0,1) lies in the I Quadrant. The tangent at any point 'P' on it meets the x-axis at 'Q'. Such that `PQ=1`. Then
The curve in parametric form is

Options

A. `x = cos theta + In tan (theta//2), y = sin theta`
B. `x = -cos theta + In tan (theta//2), y = sin theta`
C. `x = -cos theta - In tan (theta//2), y = sin theta`
D. None of these

Answer

C

Explanation

Clearly PQ = Length of tangent
`therefore" "PQ = y sqrt(1+((dx)/(dy))^(2))`
Given PQ = 1
`therefore" "y sqrt(1+((dx)/(dy))^(2)) = 1`
`therefore" "((dy)/(dx))^(2) = (y^(2))/(1-y^(2))`
`therefore" "(dy)/(dx)=(-y)/(sqrt(1-y^(2)))` (as curve lies in first quadrant)
Putting `y = sin theta`
`therefore" "cos theta(d theta)/(dx) = (-sin theta)/(cos theta)`
`therefore" "-(cos^(2)theta)/(sin theta)d theta = dx`
`therefore" "(sin theta-(1)/(sin theta))d theta = dx`
`therefore" "- cos theta - log_(e) "tan"(theta)/(2)=x + c`
`" "y = 1 "when x" = 0`
`therefore" "x = 0, theta = pi//2 therefore c = 0`
`therefore" "`The curve in parametric form is
`x = -cos theta -log_(e)"tan"((theta)/(2)), y = sin theta`