let T denote a curve `y=f(x)` which is in the first quadrant and let the point (1,0) lie on it. Let the tangent to T at a point P intersect the y-axis at `Y_(P)` and `PY_P` has length 1 for each poinit P on T. then which of the following option may be correct?

Options

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

Answer

A, B, C, D

Explanation

`(a,f(a))-=r`
`f'(x)` be differentiation of f(x) equation of tangent
`(y-f(a))=f'(a)(x-a)`
put x=0
`y-f(a)=-af'(a)`
`y=f(a)-af'(a)`
`y_(p)=(0,f(a)-af'(a))`
`py_(p)=sqrt(a^(2)+(af'(a))^(2))=1`
`a^(2)+a^(2)(f'(a))^(2)=1`
`(f'(a))^(2)=(1-a^(2))/(a^(2))`
`int(f'(x))=+-intsqrt((1-x^(2))/(x^(2)))`
put `sqrt(1-x^(2))=t`
`impliesy=+-int(-t^(2)dt)/(1-t^(2))=+-(t-(1)/(2)ln|(1+t)/(1-t)|)+c+-(t-(1)/(2)ln((1+t)^(2))/(1-t^(2)))+c=+-(sqrt(1+x^(2))+ln(1+sqrt(1+x^(2)))/(x))+c`
`implies` put x=1 and `y=0impliesc=0`