Question

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

• A. $y={\log }_e\left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}$
• B. $xy\text{'}-\sqrt{\left[1-x^2\right]}=0$
• C. $y={\log }_e\left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}$
• D. $xy\text{'}+\sqrt{\left[1-x^2\right]}=0$

Answer 1

Answer: (A), (D)

$xy\text{'}+\sqrt{\left[1-x^2\right]}=0$

Explanation for correct options:

Determining the correct option:

Draw curve $y=y\left(x\right)$ at point $\left(1,0\right)$

Let, $P\left(x,y\right),y=f\left(x\right)$

The Equation of the tangent is

$Y-y=\frac{dy}{dx}\left(X-x\right)$

$$\begin{gathered} \Rightarrow Y-y=-x\frac{dy}{dx} \\ \Rightarrow Y=y-x\frac{dy}{dx} \end{gathered}$$

So, $\left(0,y-x\frac{dy}{dx}\right)=\left(0,Y_P\right)$

From given data

$$\begin{gathered} P{Y}_P=1 \\ \sqrt{{\left(x-0\right)}^2+{\left(y-x\frac{dy}{dx}-y\right)}^2}=1 \\ \sqrt{{\left(x-0\right)}^2+{\left(x\frac{dy}{dx}\right)}^2}=1 \end{gathered}$$

Squaring on both the sides

$$\begin{gathered} x^2+{\left(x\frac{dy}{dx}\right)}^2=1 \\ {\left(x\frac{dy}{dx}\right)}^2=1-x^2 \\ \left(x\frac{dy}{dx}\right)=\pm \sqrt{1-x^2} \\ \frac{dy}{dx}=\pm \frac{\sqrt{1-x^2}}{x} \\ dy=\pm \frac{\sqrt{1-x^2}}{x}dx.............\left(1\right) \end{gathered}$$

Let,$x=\sin \theta$

On differentiating with respect to $x$,

$dx=\cos \theta d\theta$

Then,

$$\begin{gathered} dy=\pm \left(\frac{\sqrt{1-{\sin }^2\theta }}{\sin \theta }\right)\cos \theta d\theta \\ dy=\pm \left(\frac{\cos \theta }{\sin \theta }\right)\cos \theta d\theta \\ dy=\pm \left(\frac{{\cos }^2\theta }{\sin \theta }\right)d\theta \\ dy=\pm \left(\frac{1-{\sin }^2\theta }{\sin \theta }\right)d\theta \\ dy=\pm \left(\frac{1}{\sin \theta }-\frac{{\sin }^2\theta }{\sin \theta }\right)d\theta \\ dy=\pm \left(\cos ec\theta -\sin \theta \right)d\theta \end{gathered}$$

Integrating on both sides

$$\begin{gathered} y=\pm \left[\log \left|\cos ec\theta -cot\theta \right|+\cos \theta \right]+c \\ y=\pm \left[\log \left(\cos ec\theta +cot\theta \right)+\cos \theta \right]+c \end{gathered}$$

We know that,

$$\begin{gathered} \Rightarrow \sin \theta =x \\ \Rightarrow {\sin }^2\theta =x^2 \\ \Rightarrow 1-{\cos }^2\theta =x^2 \\ \Rightarrow {\cos }^2\theta =x^2-1 \\ \Rightarrow \cos \theta =\sqrt{x^2-1} \\ \therefore cot\theta =\frac{\cos \theta }{\sin \theta }=\frac{\sqrt{1-x^2}}{x}\left[\because \cos \theta =\sqrt{x^2-1}\mathrm{and}\sin \theta =x\right] \end{gathered}$$

Then,

$$\begin{gathered} y=\pm \left[-\log \left(\frac{1}{x}+\frac{\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}\right]+c \\ y=\pm \left[-\log \left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}\right]+c \end{gathered}$$

As the curve lies in the first quadrant so $y$ must be positive

$y=\left[\log \left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}\right]+c$

When,

$$\begin{gathered} \Rightarrow y=f\left(x\right),\left(1,0\right) \\ \Rightarrow \text{at}x=1,y=0 \\ \Rightarrow y\left(1\right)=0 \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow 0=\left[\log \left(\frac{1+\sqrt{1-1^2}}{1}\right)-\sqrt{1-1^2}\right]+c \\ \Rightarrow 0=0+0+c \\ \Rightarrow c=0 \end{gathered}$$

Since, $y=\left[\log \left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}\right]$

Hence, the correct answer is option (A).

Again using the negative value of equation $\left(1\right)$.

$\begin{array}{rcl}dy& =& -\frac{\sqrt{1-x^2}}{x}dx\\ & \Rightarrow & \frac{dy}{dx}=-\frac{\sqrt{1-x^2}}{x}\\ & \Rightarrow & xy\text{'}+\frac{\sqrt{1-x^2}}{x}=0\end{array}$

Hence, the correct answer is option (D).

Therefore, the correct answers are options (A) and (D).