Let T denote a curve y = y x which is in the first quadrant and let the point 1 ,   0 lie on it. Let…

1. $\mathrm{y}={\mathrm{l}\mathrm{o}\mathrm{g}}_{\mathrm{e}}\left(\frac{1+\sqrt{1-{\mathrm{x}}^2}}{\mathrm{x}}\right)-\sqrt{1-{\mathrm{x}}^2}$
2. \(x y^{\large \prime}-\sqrt{1-x^2}=0\)
3. $\mathrm{y}=-{\mathrm{l}\mathrm{o}\mathrm{g}}_{\mathrm{e}}\left(\frac{1+\sqrt{1-{\mathrm{x}}^2}}{\mathrm{x}}\right)+\sqrt{1-{\mathrm{x}}^2}$
4. \(\mathrm{xy}^{\large \prime}+\sqrt{1-\mathrm{x}^2}=0\)

Solution

Let point $\mathrm{P}$ is $(\mathrm{h},\mathrm{ }\mathrm{k})$
Now, equation of tangent at $\mathrm{P}$
$(y-k)=m_T (x- h)$ ...(i) ( $m_T=$ slope of tangent at $\mathrm{P}$ )
For $Y_P$ , put $x=0$ in.....(i)
$Y_P \left(0, k- h{m}_T\right)$
Now, as given $P{Y}_P=1$
$\sqrt{h^2+h^2{m}_T^2}=1$
$\sqrt{h^2\left(1+m_T^2\right)}=1$
$1+m_T^2=\frac{1}{h^2}$
$m_T^2=\frac{1}{h^2}-1$
$m_T^2=\frac{1-h^2}{h^2}$
$m_T=\pm \frac{\sqrt{1-h^2}}{h}$
Put $m_T=\frac{dy}{dx}\mathrm{a}\mathrm{n}\mathrm{d}h=x$
$\frac{dy}{dx}=\pm \frac{\sqrt{1-x^2}}{x}$ ...(i)
$y=\pm \left(-\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}\right)$
As the curve lies in $1^{\mathrm{s}\mathrm{t}}$ quadrant y must be positive. Hence,
$y=\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}$
Also from equation (i) only negative sign will give the correct equation of centre.
Hence,
$\frac{dy}{dx}=-\frac{\sqrt{1-x^2}}{x}$
$x{y}^{\prime }+\sqrt{1-x^2}=0$

Asked in: JEE Advanced 2019 (Paper 1)

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