What kind of hyperbola does the equation 9x^2 | vedclass

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(C) Given equation: $9x^2 - 16y^2 - 18x + 32y - 151 = 0$Rearranging terms: $9(x^2 - 2x) - 16(y^2 - 2y) = 151$Completing the square: $9(x^2 - 2x + 1) -

Vedclass · Accepted Answer

Equations of directrices: $x = \frac{21}{5}$ and $x = -\frac{11}{5}$

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(C) Given equation: $9x^2 - 16y^2 - 18x + 32y - 151 = 0$Rearranging terms: $9(x^2 - 2x) - 16(y^2 - 2y) = 151$Completing the square: $9(x^2 - 2x + 1) - 16(y^2 - 2y + 1) = 151 + 9 - 16$$9(x - 1)^2 - 16(y - 1)^2 = 144$Dividing by $144$: $\frac{(x - 1)^2}{16} - \frac{(y - 1)^2}{9} = 1$This is a hyperbola with $a^2 = 16$ $(a = 4)$ and $b^2 = 9$ $(b = 3)$.Transverse axis length $= 2a = 2(4) = 8$.Latus rectum length $= \frac{2b^2}{a} = \frac{2(9)}{4} = 4.5$.Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.Directrices are $x - 1 = \pm \frac{a}{e} = \pm \frac{4}{5/4} = \pm \frac{16}{5}$.$x = 1 \pm \frac{16}{5} \implies x = \frac{21}{5}$ or $x = -\frac{11}{5}$.Thus,option $C$ is correct.

What kind of hyperbola does the equation $9x^2 - 16y^2 - 18x + 32y - 151 = 0$ represent?

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