Differential Equations 3 Question 9
9.
If length of tangent at any point on the curve $y=f(x)$ intercepted between the point and the $X$-axis is of length $1$ . Find the equation of the curve.
(2005, 4M)
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Answer:
Correct Answer: 9. ($\sqrt{1-y^{2}}-\log \left|\frac{1+\sqrt{1-y^{2}}}{1-\sqrt{1-y^{2}}}\right|= \pm x+c$)
Solution:
- Since, the length of tangent $=\left|y \sqrt{1+\frac{d x^{2}}{d y}}\right|=1$
$ \Rightarrow \quad y^{2} (1+(\frac{d x}{d y})^{2})=1 $
$ \begin{aligned} & \therefore & \frac{d y}{d x} & = \pm \frac{y}{\sqrt{1-y^{2}}} \\ & \Rightarrow & \int \frac{\sqrt{1-y^{2}}}{y} d y & = \pm \int x d x \\ & \Rightarrow & \int \frac{\sqrt{1-y^{2}}}{y} d y & = \pm x+C \end{aligned} $
Put $y=\sin \theta \Rightarrow d y=\cos \theta d \theta$
$ \begin{aligned} & \therefore \quad \int \frac{\cos \theta}{\sin \theta} \cdot \cos \theta d \theta= \pm x+C \\ & \Rightarrow \quad \int \frac{\cos ^{2} \theta}{\sin ^{2} \theta} \cdot \sin \theta d \theta= \pm x+C \end{aligned} $
Again put $\cos \theta=t \Rightarrow-\sin \theta d \theta=d t$
$ \begin{aligned} & \therefore \quad-\int \frac{t^{2}}{1-t^{2}} d t= \pm x+C \\ & \Rightarrow \quad \int 1-\frac{1}{1-t^{2}} d t= \pm x+C \\ & \Rightarrow \quad 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 \end{aligned} $
