Question

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

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

Answer 1

Answer: (A), (B)

The correct options are
A $\sqrt{1-y^2}-\log \left|\frac{1+\sqrt{1-y^2}}{1-\sqrt{1-y^2}}\right|=x+c$
B $\sqrt{1-y^2}-\log \left|\frac{1+\sqrt{1-y^2}}{1-\sqrt{1-y^2}}\right|=-x+c$

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{dy}{dx}=\pm \frac{y}{\sqrt{1-y^2}}$$\Rightarrow \int \frac{\sqrt{1-y^2}}{y}dy=\pm \int xdx$

$\Rightarrow \int \frac{\sqrt{1-y^2}}{y}dy=\pm x+c$

Substitute $y=\sin \theta \Rightarrow dy=\cos \theta d\theta$

$\therefore \int \frac{\cos \theta }{\sin \theta }.\cos \theta d\theta =\pm x+c\Rightarrow \int \frac{{\cos }^2\theta }{{\sin }^2\theta }.\sin \theta d\theta =\pm x+c$

Again substitute $\cos \theta =t\Rightarrow -\sin \theta d\theta =dt$

$\therefore \int \frac{t^2}{1-t^2}dt=\pm x+c,\Rightarrow \int (1-\frac{1}{1-t^2})dt=\pm x+c$

$\Rightarrow t-\log \left|\frac{1+t}{1-t}\right|=\pm x+c\Rightarrow \sqrt{1-y^2}-\log \left|\frac{1+\sqrt{1-y^2}}{1-\sqrt{1-y^2}}\right|=\pm x+c$