Let $\Gamma$ denote a curve $y=y(x)$ which is in the first quadrant and let the point $(1,0)$ lies o

Question

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

Professor PC Thomas Classes · Answer

(b, d) : Equation of tangent is\ $Y-y=y^{\prime}(X-x)$\ Then, $Y_{P}=\left(0, y-x y^{\prime}ight)$\ $\left[\becauseight.$ It intersects the $y$-axis at $\left.Y_{P}ight]$\ Since $P Y_{P}$ has length 1 for each point $P(x, y)$\ on Г. $	herefore x^{2}+\left(x y^{\prime}ight)^{2}=1$\ $\Rightarrow \frac{d y}{d x}= \pm \sqrt{\frac{1-x^{2}}{x^{2}}}$\ $\Rightarrow \int d y= \pm \int \frac{\sqrt{1-x^{2}}}{x} d x$\ Put $x=\sin 	heta \Rightarrow d x=\cos 	heta \mathrm{d} 	heta$\ $	herefore y= \pm \int \frac{\cos ^{2} 	heta}{\sin 	heta} d 	heta= \pm \int\left(\frac{1-\sin ^{2} 	heta}{\sin 	heta}ight) d 	heta$\ $= \pm[\ln (\operatorname{cosec} 	heta-\cot 	heta)+\cos 	heta]+c$\ $\Rightarrow y= \pm\left[\log _{e}\left|\frac{1-\sqrt{1-x^{2}}}{x}ight|+\sqrt{1-x^{2}}ight]+c$\ $\Rightarrow y= \pm\left[\sqrt{1-x^{2}}-\log _{e}\left|\frac{1+\sqrt{1-x^{2}}}{x}ight|ight]+c$\ As $y(1)=0$, so we have $c=0$\ $	herefore y= \pm\left[-\log _{e}\left(\frac{1+\sqrt{1-x^{2}}}{x}ight)+\sqrt{1-x^{2}}ight]$\ $\Rightarrow \quad y=\log _{e}\left(\frac{1+\sqrt{1-x^{2}}}{x}ight)-\sqrt{1-x^{2}}$\ [As given curve lies in first quadrant]\ Also, $\frac{d y}{d x}=\frac{-\sqrt{1-x^{2}}}{x}$\