The eccentricity of the hyperbola 16x^{2} - 3 | vedclass

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Question

(A) Given equation: $16x^{2} - 3y^{2} - 32x - 12y - 44 = 0$ Rearranging terms: $16(x^{2} - 2x) - 3(y^{2} + 4y) = 44$ Completing the square: $16(x^{2}

Vedclass · Accepted Answer

$\sqrt{\frac{19}{3}}$

Vedclass · Answer

(A) Given equation: $16x^{2} - 3y^{2} - 32x - 12y - 44 = 0$ Rearranging terms: $16(x^{2} - 2x) - 3(y^{2} + 4y) = 44$ Completing the square: $16(x^{2} - 2x + 1) - 3(y^{2} + 4y + 4) = 44 + 16 - 12$ $16(x - 1)^{2} - 3(y + 2)^{2} = 48$ Dividing by $48$: $\frac{(x - 1)^{2}}{3} - \frac{(y + 2)^{2}}{16} = 1$ Comparing with standard form $\frac{(x - h)^{2}}{a^{2}} - \frac{(y - k)^{2}}{b^{2}} = 1$,we get $a^{2} = 3$ and $b^{2} = 16$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$. $e = \sqrt{1 + \frac{16}{3}} = \sqrt{\frac{3 + 16}{3}} = \sqrt{\frac{19}{3}}$.

The eccentricity of the hyperbola $16x^{2} - 3y^{2} - 32x - 12y - 44 = 0$ is

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