Let `lceiling` denote a curve `y=y(x)` which is in the first quadrant and let the point `(1,0)` lie on it . Let the tangant ot `lceiling` at a point P intersect the y-axis at `Y_P`. If `PY_P` has length l for each point P on `lceiling`, then which of the following options is/ are correct ?

Options

A. `xy'+sqrt(1-x^2)=0`
B. `xy'-sqrt(1-x^2)=0`
C. `y=log_(e)((1+sqrt(1-x^2))/(x))-sqrt(1-x^2)`
D. `y=-log_e((1+sqrt(1-x^2))/(x))+sqrt(1-x^2)`

Answer

A, C

Explanation

Let a point P(h,k)on the curve `y=y(x)`, so equation of tangent to the curve at point P is
`y-k=((dy)/(dx))_(h,k) (x-h)` ....(i)
Now, the tangent (i) intersect the Y-axis at `Y_p` , so coordinates `Y_(p)` is `(0,k-h(dy)/(dx))`, where `(dy)/(dx) =((dy)/(dx))_(h,k)`
So, `PY_p=1`(given)
`rArr sqrt(h^2+h^2((dy)/(dx))^2)=1`
`rArr (dy)/(dx)=+-(sqrt(1-x_1^2))/(x)" "["on replacing h by x"]`
`rArr dy= +-sqrt(1-x^2)/(x)dx`
On putting `x=sintheta, dx =cos thetad theta`,we get
`dy =+-sqrt(1-sin^2)/(sin theta)cos theta d theta =+-(cos^2 theta)/(sin theta)d theta`
`=+-(cosec theta -sintheta )d theta`
`rArr y=+-[In (cosec theta -cot theta )+cos theta]+C`
` y=+-[-In ((1+sqrt((1-x^2)))/(x))+sqrt(1-sin^2)]+C`
`rArr y+-[In ((1-costheta)/(sin theta))+cos theta]+C`
`rArr y=+-[In (1-sqrt(1-x^2)/(x))+sqrt(1-x^2)]+C " "[because x=sintheta]`
` =+-[-In (1+sqrt(1-x^2))/(x)+sqrt(1-x^2)]+C` [On ratioalization]
`because` The curve is in the first quadrant so y must be positive, so
`y=In(1+sqrt((1-x^2))/(x))-sqrt(1-x^2)+C`
As curve passes through `(1,0)`, so
`0=0-0+c rArr c=0`, so required curve is
`y=In (1+sqrt(1-x^2))/(x)=sqrt(1-x^2)`
and required differential equation is
`(dy)/(dx)=-(sqrt(1-x^2))/(x)`
`rArr xy'+sqrt(1-x^2)=0`
Hence, options (a) and (c) are correct.