Find the orthogonal trajectory of x² + cy² = 1

1.	Find the orthogonal trajectory of x² + cy² = 1
Answer» $Given \:x^{2} + cy^{2} = 1$ $\implies cy^{2} = 1 - x^{2}$ $\implies c = \frac{1 - x^{2}}{y^{2}}$ /* Do the DIFFERENTIATION both SIDES , we GET */ $\implies 0 = \frac{y^{2} ( -2x) - (1-x^{2})\times 2y\frac{dy}{dx}}{y^{4}}$ $\implies 0 = -2xy^{2} + 2y(x^{2}-1)\frac{dy}{dx}$ $\implies 2xy^{2} = 2y(x^{2}-1)\frac{dy}{dx}$ $\implies \frac{xy}{x^{2}-1} = \frac{dy}{dx}$ $Differential \:equation \:of\\Orthogonal \: trajectory$ $-\frac{dx}{dy} = \frac{xy}{x^{2}-1}$ $\implies - \int \frac{x^{2}-1}{x} = \int y dy$ $\implies - \int \Big( x - \frac{1}{x}\Big) dx = \int y dy$ $\implies -x + log x = \frac{y^{2}}{2} + c$ $\green { This \:is \:\:equation\:of \: Orthogonal}\\\green { trajectory \:of \: given \: family\:of \:curves}$ •••♪

Answer»

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