Question

If length of tangent at any point on the curve $y=f\left(x\right)$ intercepted between the point and the xaxis is of
length 1. Find the equation of the curve.

Answer 1

Since, the length of tangent $=\left|y\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}\right|=1$ $\Rightarrow y^2(1+{\left(\frac{dx}{dy}\right)}^2)=1$
$\therefore \frac{dx}{dy}=\frac{y}{\sqrt{1-y^2}}$
$\Rightarrow \int \frac{\sqrt{1-y^2}}{y}dy=\pm \int xdx\Rightarrow \int \frac{\sqrt{1-y^2}}{y}dy=\pm x+c$
Put $y=\sin \theta$ $\Rightarrow dy=\cos \theta d\theta$
$\therefore \int \frac{\cos \theta }{\sin \theta }.\cos \theta d\theta =\pm x+c\Rightarrow \int \frac{{\cos }^2\theta }{{\sin }^2\theta }.\sin \theta d\theta =\pm x+c$
Again put $\cos \theta =t\Rightarrow -\sin \theta d\theta =dt$
$\therefore -\int \frac{t^2}{1-t^2}dt=\pm x+c,$
$\Rightarrow \int (1-\frac{1}{1-t^2})dt=\pm x+c\Rightarrow t-\log \left|\frac{1+t}{1-t}\right|=\pm x+c$
$\Rightarrow \sqrt{1-y^2}-\log \left|\frac{1+\sqrt{1-y^2}}{1-\sqrt{1-y^2}}\right|=\pm x+c$
$\Rightarrow f\left(x\right)=e^C{e}^x$
$\because f\left(0\right)=0\Rightarrow e^C=0$, a contradiction
$\therefore f\left(x\right)=0\forall x\in R\Rightarrow f\left(ln5\right)=0$
and $f^{\prime }\left(x\right)=f\left(x\right)$
if $f\left(x\right)\ne 0$
$\Rightarrow \frac{f^{{}^{\prime }}\left(x\right)}{f\left(x\right)}=1\Rightarrow \ln f\left(x\right)=x+c$ $\Rightarrow f\left(x\right)=e^c{e}^x$
$\because f\left(0\right)=0\Rightarrow e^c=0$, a contradiction
$\therefore f\left(x\right)=0\forall x\in R\Rightarrow f\left(\ln 5\right)=0$