Question

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 $P{Y}_P$ has length 1 for each point $P$ on $\Gamma$, then which of the following options is/are correct?

• A. $y=lo{g}_e\left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}$
• B. $x{y}^{\prime }=\sqrt{1-x^2}=0$
• C. $y=-lo{g}_e\left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}$
• D. $x{y}^{\prime }-\sqrt{1-x^2}=0$

Answer 1

Answer: (A), (B)

Let point $P(x,y)$
Equation of tangent at $P$;
$Y-y=\frac{dy}{dx}\left(X-x\right)$
So $Y_P\left(0,y-x\frac{dy}{dx}\right)$
$P{Y}_P=\sqrt{x^2+{\left(x\frac{dy}{dx}\right)}^2}=1$
$\Rightarrow \frac{dy}{dx}=\pm \frac{\sqrt{1-x^2}}{x}$
$\Rightarrow dy=\pm \left(\frac{\sqrt{1-x^2}}{x}\right)dxletx=sin\theta ,dx=cos\theta d\theta$
$\Rightarrow dy=\pm \left(cosec\theta -sin\theta \right)d\theta$
$\Rightarrow y=\pm \left[-In\left(cosec\theta -cot\theta \right)+cos\theta \right]+C$
$\Rightarrow y=\pm \left[-In\left(cosec\theta +cot\theta \right)+cos\theta \right]+C$
$y=\pm \left[-In\left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}\right]+C$
As the curve lies in the $1^{st}$ quadrant so y must be positive
$y=In\left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}+C$
$\because y\left(1\right)=0\Rightarrow C=0$
$\Rightarrow y=In\left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}$
Also the correct differential equation will be
$\frac{dy}{dx}=-\frac{\sqrt{1-x^2}}{x}$
$\Rightarrow x{y}^{\prime }+\sqrt{1-x^2}=0$