The foci of the hyperbola 5x^2 - 6y^2 - 10x - | vedclass

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Question

(D) Given equation: $5x^2 - 6y^2 - 10x - 24y - 34 = 0$ Rearranging terms: $5(x^2 - 2x) - 6(y^2 + 4y) = 34$ Completing the square: $5(x^2 - 2x + 1) - 6

Vedclass · Accepted Answer

$\left(1 \pm \sqrt{\frac{11}{2}}, -2\right)$

Vedclass · Answer

(D) Given equation: $5x^2 - 6y^2 - 10x - 24y - 34 = 0$ Rearranging terms: $5(x^2 - 2x) - 6(y^2 + 4y) = 34$ Completing the square: $5(x^2 - 2x + 1) - 6(y^2 + 4y + 4) = 34 + 5 - 24$ $5(x - 1)^2 - 6(y + 2)^2 = 15$ Dividing by $15$: $\frac{(x - 1)^2}{3} - \frac{(y + 2)^2}{2.5} = 1$ Here,$a^2 = 3$ and $b^2 = 2.5 = \frac{5}{2}$. Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5/2}{3}} = \sqrt{1 + \frac{5}{6}} = \sqrt{\frac{11}{6}}$. Foci are given by $(h \pm ae, k)$,where $(h, k) = (1, -2)$. $ae = \sqrt{3} 	imes \sqrt{\frac{11}{6}} = \sqrt{\frac{33}{6}} = \sqrt{\frac{11}{2}}$. Thus,the foci are $\left(1 \pm \sqrt{\frac{11}{2}}, -2ight)$.

The foci of the hyperbola $5x^2 - 6y^2 - 10x - 24y - 34 = 0$ are

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