SATHEE: Differential Equations 3 Question 9

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. ( $\sqrt{1 - y^{2}} - \log | \frac{1 + \sqrt{1 - y^{2}}}{1 - \sqrt{1 - y^{2}}} | = \pm x + c$ )

Solution:

Since, the length of tangent $= | y \sqrt{1 + \frac{d x^{2}}{d y}} | = 1$

$\Rightarrow y^{2} (1 + (\frac{d x}{d y})^{2}) = 1$

$\begin{aligned} ∴ & \frac{d y}{d x} & = \pm \frac{y}{\sqrt{1 - y^{2}}} \\ \Rightarrow & \int \frac{\sqrt{1 - y^{2}}}{y} d y & = \pm \int x d x \\ \Rightarrow & \int \frac{\sqrt{1 - y^{2}}}{y} d y & = \pm x + C \end{aligned}$

Put $y = \sin θ \Rightarrow d y = \cos θ d θ$

$\begin{aligned} ∴ \int \frac{\cos θ}{\sin θ} \cdot \cos θ d θ = \pm x + C \\ \Rightarrow \int \frac{\cos^{2} θ}{\sin^{2} θ} \cdot \sin θ d θ = \pm x + C \end{aligned}$

Again put $\cos θ = t \Rightarrow - \sin θ d θ = d t$

$\begin{aligned} ∴ - \int \frac{t^{2}}{1 - t^{2}} d t = \pm x + C \\ \Rightarrow \int 1 - \frac{1}{1 - t^{2}} d t = \pm x + C \\ \Rightarrow t - \log | \frac{1 + t}{1 - t} | = \pm x + C \\ \Rightarrow \sqrt{1 - y^{2}} - \log | \frac{1 + \sqrt{1 - y^{2}}}{1 - \sqrt{1 - y^{2}}} | = \pm x + C \end{aligned}$

Differential Equations 3 Question 9

9.

Answer:

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Differential Equations 3 Question 9

9.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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