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

$\Gamma: y = y(x) \text{ in first quadrant}, \quad (1,0) \in \Gamma$

$\text{Tangent at } P \text{ meets } y\text{-axis at } Y_P; \quad |P Y_P| = 1$

$\text{To find: which of (1)--(4) are correct}$

$\text{Tangent at } P(x,y):\ Y - y = y'(X - x)$

$\Rightarrow Y_P = (0,\ y - x y')$

$PY_P^2 = x^2 + (x y')^2 = x^2(1 + y'^2) = 1$

$\Rightarrow x y' = \pm\sqrt{1 - x^2}$

$y > 0,\ y(1) = 0 \Rightarrow y' < 0 \Rightarrow x y' = -\sqrt{1 - x^2}$

$\Rightarrow x y' + \sqrt{1 - x^2} = 0 \Rightarrow (4)\ \text{TRUE}$

$y' = -\dfrac{\sqrt{1-x^2}}{x}$

$y = \ln\!\left(\dfrac{1+\sqrt{1-x^2}}{x}\right) - \sqrt{1-x^2} + C$

$x = 1,\ y = 0 \Rightarrow C = 0$

$\Rightarrow y = \ln\!\left(\dfrac{1+\sqrt{1-x^2}}{x}\right) - \sqrt{1-x^2} \Rightarrow (3)\ \text{TRUE}$

$(1)\ y = -(\ldots) < 0 \Rightarrow \text{not first quadrant}$

$(2)\ \text{wrong sign of } y' \Rightarrow \text{incorrect}$

$\text{Correct: } (3)\ \text{and}\ (4)$

$\boxed{(3)\ \text{and}\ (4)}$