SATHEE: Differential Equations 3 Question 13

Differential Equations 3 Question 13

13. A curve passing through the point $(1, 1)$ has the property that the perpendicular distance of the origin from the normal at any point $P$ of the curve is equal to the distance of $P$ from the $X$ -axis. Determine the equation of the curve.

(1999, 10M)

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Answer:

Correct Answer: 13. $\frac{1}{3} \tan^{- 1} \frac{4}{5} \tan 4 x + \tan^{- 1} \frac{3}{4} - \frac{3}{5} - \frac{5 x}{3}$

Solution:

Equation of normal at point $(x, y)$ is

$Y - y = - \frac{d x}{d y} (X - x)$

Distance of perpendicular from the origin to Eq. (i)

$= \frac{| y + \frac{d x}{d y} \cdot x |}{\sqrt{1 + \frac{d x}{d y}}}$

Also, distance between $P$ and $X$ -axis is $| y |$ .

$\begin{aligned} ∴ \frac{| y + \frac{d x}{d y} \cdot x |}{\sqrt{1 + \frac{d x}{d y}}} = | y | \\ \Rightarrow y^{2} + \frac{d x}{d y} \cdot x^{2} + 2 x y \frac{d x}{d y} = y^{2} 1 + \frac{d x^{2}}{d y} \\ \Rightarrow \frac{d x^{2}}{d y} (x^{2} - y^{2}) + 2 x y \frac{d x}{d y} = 0 \\ \Rightarrow \frac{d x}{d y} \frac{d x}{d y} (x^{2} - y^{2}) + 2 x y = 0 \\ \Rightarrow \frac{d x}{d y} = 0 or \frac{d y}{d x} = \frac{y^{2} - x^{2}}{2 x y} \\ But \frac{d x}{d y} = 0 \end{aligned}$

Since, curve passes through $(1, 1)$ , we get the equation of the curve as $x = 1$ .

The equation $\frac{d y}{d x} = \frac{y^{2} - x^{2}}{2 x y}$ is a homogeneous equation.

Put $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

$v + x \frac{d v}{d x} = \frac{v^{2} x^{2} - x^{2}}{2 x^{2} v}$

$\Rightarrow x \frac{d v}{d x} = \frac{v^{2} - 1}{2 v} - v = \frac{v^{2} - 1 - 2 v^{2}}{2 v} = - \frac{v^{2} + 1}{2 v}$

$\Rightarrow \frac{- 2 v}{v^{2} + 1} d v = \frac{d x}{x}$

$\Rightarrow c_{1} - \log (v^{2} + 1) = \log | x |$

$\Rightarrow \log | x | (v^{2} + 1) = c_{1} \Rightarrow | x | \frac{y^{2}}{x^{2}} + 1 = e^{c_{1}}$

$\Rightarrow x^{2} + y^{2} = \pm e^{c_{1}} x$ or $x^{2} + y^{2} = \pm e^{c} x$ is passing through $(1, 1)$ .

$\begin{aligned} ∴ & 1 + 1 & = \pm e^{c} \cdot 1 \\ \Rightarrow & \pm e^{c} & = 2 \end{aligned}$

Hence, required curve is $x^{2} + y^{2} = 2 x$ .

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Differential Equations 3 Question 13

13. A curve passing through the point $(1, 1)$ has the property that the perpendicular distance of the origin from the normal at any point $P$ of the curve is equal to the distance of $P$ from the $X$ -axis. Determine the equation of the curve.

Answer:

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Differential Equations 3 Question 13

13. A curve passing through the point (1,1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the X-axis. Determine the equation of the curve.

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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13. A curve passing through the point $(1, 1)$ has the property that the perpendicular distance of the origin from the normal at any point $P$ of the curve is equal to the distance of $P$ from the $X$ -axis. Determine the equation of the curve.