Question

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

• A. $\sqrt{1-y^2}+ln|(1-\sqrt{1-y^2})/y|=\pm x+c$
• B. $\sqrt{1-y^2}-ln|(1-\sqrt{1-y^2})/y|=\pm x+c$
• C. $\sqrt{1-y^2}+ln|(1+\sqrt{1-y^2})/y|=\pm x+c$
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A $\sqrt{1-y^2}+ln|(1-\sqrt{1-y^2})/y|=\pm x+c$
C $\sqrt{1-y^2}+ln|(1+\sqrt{1-y^2})/y|=\pm x+c$

Let point P(x,y) be the point lying on curve at which tangent is drawn. Let this tangent intersect x-axis at $\theta$.

Now, Equation of $P\theta \Rightarrow (Y-y)=\frac{dy}{dx}(X-x)$

Now, for $\theta ,Y=0$

$X=x-y\frac{dx}{dy}$

Therefore, $P\equiv (x,y)$ ; $\theta =(x-y\frac{dx}{dy},0)$

Now,$P\theta =l$

$\sqrt{(x-x+y\frac{dx}{dy}{)}^2+(y-0{)}^2}=l$

$y^2(\frac{dx}{dy}{)}^2+y^2=l^2$

$\frac{dx}{dy}=\pm \frac{\sqrt{l^2-y^2}}{y}$

$dx=\pm \frac{\sqrt{l^2-y^2}}{y}dy$

Now, integrating on both sides, we get

$\int dx=\int \pm \frac{\sqrt{l^2-y^2}}{y}dy$

In RHS, put $y=l\sin \theta$

$\Rightarrow dy=l\cos \theta d\theta$

$x+c=\int \pm \left(\frac{\sqrt{l^2-l^2{\sin }^2\theta }}{l\sin \theta }\right)$ $l\cos \theta d\theta$

$x+c=\int l\frac{{\cos }^2\theta }{\sin \theta }d\theta$

$x+c=l\int \frac{1-{\sin }^2\theta }{\sin \theta }d\theta =l\int (\csc \theta -\sin \theta )d\theta$

$x+c=l\left[\ln \right(\csc \theta -\cot \theta )+\cos \theta ]$

as $y=l\sin \theta$

$\Rightarrow \csc \theta =\frac{l}{y}$, $\cos \theta =\sqrt{1-\frac{y^2}{l^2}}$

$\cot \theta =\frac{\sqrt{l^2-y^2}}{y}$

Therefore,

$x+c=\pm l\left[\ln \right(\frac{l}{y}-\frac{\sqrt{l^2-y^2}}{y})+\frac{\sqrt{l^2-y^2}}{l}$

Now, taking $l=1$

$\sqrt{1-y^2}\pm \ln \mid \frac{(1-\sqrt{1-y^2})}{y}\mid =\pm x+c$