Question

Let $\tau$ denote a curve $y=y\left(x\right)$ which is in the first quadrant and let the point $(1,0)$ lie on it. Let the tangent to $\tau$ at a point $P$ intersect the y-axis at $Y_p$. If $P{Y}_p$ has length $1$ for each point $P$ on $\tau$, then which of the following options is /are correct?

• A. $y={\log }_e\frac{1+\sqrt{1-x^2}}{x}-\sqrt{1-x^2}$
• B. $x{y}^{\prime }+\sqrt{1-x^2}=0$
• C. $y=-{\log }_e\frac{1+\sqrt{1-x^2}}{x}+\sqrt{1-x^2}$
• D. $x{y}^{\prime }-\sqrt{1-x^2}=0$

Answer 1

Answer: (A), (B)

The correct options are
A $y={\log }_e\frac{1+\sqrt{1-x^2}}{x}-\sqrt{1-x^2}$
B $x{y}^{\prime }+\sqrt{1-x^2}=0$

Equation of tangent,
$(y-y_1)=\frac{d{y}_1}{d{x}_1}(x-x_1)$
Coordinate of $Y_p(0,y_1-x_1\frac{d{y}_1}{d{x}_1})$
Distance of $P{Y}_p=\sqrt{(x_1{)}^2+(y_1-y_1+x_1\frac{d{y}_1}{d{x}_1}{)}^2)}=1$
Generalising slope of line
$\frac{dy}{dx}=\pm \frac{\sqrt{1-x^2}}{x}\Rightarrow x{y}^{\prime }\mp \sqrt{1-x^2}=0$
$dy=\pm \frac{\sqrt{1-x^2}}{x}dx$
$\because \frac{dy}{dx}$ cannot be positive i.e $f\left(x\right)$ cannot be increasing in first quadrant $\forall x\in (0,1)$
Intregrating both side
$y=-\int \frac{\sqrt{1-x^2}}{x}dx$
Let $1-x^2=t^2\Rightarrow -xdx=tdt$
$\Rightarrow y=-\int \frac{t^2}{t^2-1}dt$
$\Rightarrow y=-\int 1dt+\int \frac{dt}{1-t^2}dx$
$\Rightarrow y=-t+\frac{1}{2}{\log }_e|\frac{1+t}{1-t}|+c$
$\Rightarrow y=-\sqrt{1-x^2}+\frac{1}{2}{\log }_e|\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}|+c$
Point $(1,0)$ lies on curve $\tau ,\therefore c=0$
$y=-\sqrt{1-x^2}+\frac{1}{2}{\log }_e|\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}|$
$y={\log }_e\frac{1+\sqrt{1-x^2}}{x}-\sqrt{1-x^2}$