Question

find the orthogonal trajectory of
x² + cy² = 1 ​

Answer 1

$Given \:x^{2} + cy^{2} = 1$

$\implies cy^{2} = 1 - x^{2}$

$\implies c = \frac{1 - x^{2}}{y^{2}}$

/* Do the Differentiation both sides , we get */

$\implies 0 = \frac{y^{2} ( -2x) - (1-x^{2})\times 2y\frac{dy}{dx}}{y^{4}}$

$\implies 0 = -2xy^{2} + 2y(x^{2}-1)\frac{dy}{dx}$

$\implies 2xy^{2} = 2y(x^{2}-1)\frac{dy}{dx}$

$\implies \frac{xy}{x^{2}-1} = \frac{dy}{dx}$

$Differential \:equation \:of\\Orthogonal \: trajectory$

$-\frac{dx}{dy} = \frac{xy}{x^{2}-1}$

$\implies - \int \frac{x^{2}-1}{x} = \int y dy$

$\implies - \int \Big( x - \frac{1}{x}\Big) dx = \int y dy$

$\implies -x + log x = \frac{y^{2}}{2} + c$

$\green { This \:is \:\:equation\:of \: Orthogonal}\\\green { trajectory \:of \: given \: family\:of \:curves}$

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Answer 2

Answer:

hi

Step-by-step explanation: