Question

If length of tangent at any point on the curve $y=f\left(x\right)$ intercepted between the point and the $x-$ axis is of length $1$ units, then the equation of the curve is:

• A. $\ln \left|\frac{1-\sqrt{1-y^2}}{y}\right|+\sqrt{1-x^2}=\pm x+c$
• B. $\ln \left|\frac{1-\sqrt{1-x^2}}{x}\right|+\sqrt{1-y^2}=\pm y+c$
• C. $\ln \left|\frac{1-\sqrt{1-y^2}}{y}\right|+\sqrt{1-y^2}=\pm x+c$
• D. $\ln \left|\frac{1-\sqrt{1-x^2}}{x}\right|+\sqrt{1-x^2}=\pm y+c$

Answer 1

The correct option is C $\ln \left|\frac{1-\sqrt{1-y^2}}{y}\right|+\sqrt{1-y^2}=\pm x+c$

We know that length of tangent to curve $y=f\left(x\right)$ is given by
$\left|\frac{y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}{\left(\frac{dy}{dx}\right)}\right|$
As per question $\left|\frac{y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}{\left(\frac{dy}{dx}\right)}\right|=1$
$\Rightarrow y^2(1+{\left(\frac{dy}{dx}\right)}^2)={\left(\frac{dy}{dx}\right)}^2$
$\Rightarrow {\left(\frac{dy}{dx}\right)}^2=\frac{y^2}{1-y^2}\Rightarrow \frac{dy}{dx}=\pm \frac{y}{\sqrt{1-y^2}}$
$\Rightarrow \int \frac{\sqrt{1-y^2}}{y}dy=\int \pm dx$
Put $y=\sin \theta$ so that $dy=\cos \theta d\theta$
$\Rightarrow \int \frac{\cos \theta }{\sin \theta }\cos \theta d\theta =\pm x+c$
$\Rightarrow \int (\text{cosec}\theta -\sin \theta )d\theta =\pm x+c$
$\Rightarrow \ln |\text{cosec}\theta -\cot \theta |+\cos \theta =\pm x+c$
$\Rightarrow \ln \left|\frac{1-\sqrt{1-y^2}}{y}\right|+\sqrt{1-y^2}=\pm x+c$