JEE Advanced 2019 — Paper 1 — Question 35

Let $P=(x,y)$ on $\Gamma$. The tangent at $P$ is $Y - y = y'(X-x)$. Its $y$-intercept is $Y_P = (0, y - x y')$. Distance $PY_P = \sqrt{(x-0)^2 + (y-(y-xy'))^2} = \sqrt{x^2 + (xy')^2} = 1$. Thus $x^2(1+(y')^2)=1 \Rightarrow (y')^2 = \frac{1-x^2}{x^2} \Rightarrow y' = \pm \frac{\sqrt{1-x^2}}{x}$. The curve lies in the first quadrant and passes through $(1,0)$.

For the positive sign, $y'(x)$ is positive for $0<x<1$, so $y$ decreases as $x$ moves left from 1, giving negative $y$ values; this contradicts the quadrant condition.

Hence we take the negative sign: $y' = -\frac{\sqrt{1-x^2}}{x}$ which gives $x y' + \sqrt{1-x^2} = 0$ (option B). Separate variables: $dy = -\frac{\sqrt{1-x^2}}{x}dx$. Let $t = \sqrt{1-x^2}$. Then $x dx = -t dt$ and $\frac{\sqrt{1-x^2}}{x}dx = -\frac{t^2}{1-t^2}dt$. Integrate: $\int \frac{\sqrt{1-x^2}}{x} dx = \int -\frac{t^2}{1-t^2} dt = \int \left(1+\frac{1}{t^2-1}\right) dt = t + \frac12 \ln\left|\frac{t-1}{t+1}\right| + C$. For $0\le t<1$, $\left|\frac{t-1}{t+1}\right| = \frac{1-t}{1+t}$.

Hence $\frac12 \ln\frac{1-t}{1+t} = - \ln\frac{1+t}{\sqrt{1-t^2}} = -\ln\frac{1+\sqrt{1-x^2}}{x}$. Thus $\int \frac{\sqrt{1-x^2}}{x} dx = \sqrt{1-x^2} - \ln\frac{1+\sqrt{1-x^2}}{x} + C$. Therefore $y = -\left(\sqrt{1-x^2} - \ln\frac{1+\sqrt{1-x^2}}{x} + C\right) = \ln\frac{1+\sqrt{1-x^2}}{x} - \sqrt{1-x^2} - C$. Condition $y(1)=0$ gives $0 = \ln\frac{1+0}{1} - 0 - C \Rightarrow C=0$. Hence $y = \ln\frac{1+\sqrt{1-x^2}}{x} - \sqrt{1-x^2}$ (option A). So options A and B are correct.