Let B denote a curve $y=y\left(x\right)$ which is in the first quadrant and let the point $\left(1,0\right)$  lie on it. Let the tangent to B at a point P intersect the $y-axis$  at $Y_P$

 If $P{Y}_P$  has length 1 for each point P on B, then which of the following options is/are correct?

 Equation of tangent at $P\left(x,y\right)$  is  $Y-y$= $\frac{dy}{dx}\left(X-x\right)$  $\Rightarrow$  $xdy-ydx=xdy-ydx$ .So $Y_p=y-\frac{xdy}{dx}$  $\left(\sin ce X=0\right)$

$$\begin{gathered} \begin{array}{cc}Dis\tan ce between P\left(x,y\right) and Y_P\left(0,y-x\frac{dy}{dx}\right)=P{Y}_P=\sqrt{x^2+x^2{\left(\frac{dy}{dx}\right)}^2}=1 & \\ \\ \Rightarrow \frac{dy}{dx}=\pm \frac{\sqrt{1-x^2}}{x}& \\ \text{ So }\int dy=\pm \int \frac{\sqrt{1-x^2}}{x}dx& \\ =\pm \left[\int \frac{1-x^2}{x\sqrt{1-x^2}}dx\right]& \\ =\pm \left[\int \frac{xdx}{x^2\sqrt{1-x^2}}-\int \frac{xdx}{\sqrt{1-x^2}}\right] & \\ \text{ Let }1-x^2=t^2\Rightarrow xdx=-tdt& \\ \text{ So }y=\pm \left[-\int \frac{tdt}{\left(1-t^2\right)t}+\int \frac{tdt}{t}\right] & \\ =\pm \left[\frac{1}{2}\ln \left|\frac{t-1}{t+1}\right|+t\right]& \\ y=-\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}+c& \\ y=\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}+C& \end{array} \end{gathered}$$

       put x=1 and y=0   $then c=0$ but as function lies in 1st quadrant so $y=\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}$ & $\frac{dy}{dx}=-\frac{\sqrt{1-x^2}}{x}$