The equations of the transverse and conjugate | vedclass

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(C) Given equation: $16x^2 - y^2 + 64x + 4y + 44 = 0$Rearranging terms: $16(x^2 + 4x) - (y^2 - 4y) = -44$Completing the square: $16(x^2 + 4x + 4) - (y

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$y = 2, x + 2 = 0$

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(C) Given equation: $16x^2 - y^2 + 64x + 4y + 44 = 0$Rearranging terms: $16(x^2 + 4x) - (y^2 - 4y) = -44$Completing the square: $16(x^2 + 4x + 4) - (y^2 - 4y + 4) = -44 + 64 - 4$$16(x + 2)^2 - (y - 2)^2 = 16$Dividing by $16$: $\frac{(x + 2)^2}{1} - \frac{(y - 2)^2}{16} = 1$This is a horizontal hyperbola with center $(-2, 2)$.The transverse axis is parallel to the $x$-axis,given by $y = 2$.The conjugate axis is parallel to the $y$-axis,given by $x = -2$ (or $x + 2 = 0$).

The equations of the transverse and conjugate axes of the hyperbola $16x^2 - y^2 + 64x + 4y + 44 = 0$ are

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