How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola $4 x^{2} - 25 y^{2} - 8 x - 96 = 0$ $4x^2-25y^2-8x-96=0$ ?

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How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola $4 x^{2} - 25 y^{2} - 8 x - 96 = 0$ $4x^2-25y^2-8x-96=0$ ?

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Answer 1 · 2016-10-27T05:10:23

The vertices are $(6, 0)$ $(6,0)$ and $(- 4, 0)$ $(-4,0)$
The equations of the asymptotes are $y = \frac{2}{5} (x - 1)$ $y=2/5(x-1)$ and $y = - \frac{2}{5} (x - 1)$ $y=-2/5(x-1)$
The foci are $(1 + \sqrt{29}, 0)$ $(1+sqrt29,0)$ and $(1 - \sqrt{29}, 0)$ $(1-sqrt29,0)$

Explanation:

The equation is $4 x^{2} - 8 x - 25 y^{2} = 96$ $4x^2-8x-25y^2=96$

Completing the square,give

$4 (x^{2} - 2 x) - 25 y^{2} = 96$ $4(x^2-2x)-25y^2=96$

$4 (x^{2} - 2 x + 1) - 25 y^{2} = 96 + 4$ $4(x^2-2x+1)-25y^2=96+4$

$4 {(x - 1)}^{2} - 25 y^{2} = 100$ $4(x-1)^2-25y^2=100$
Dividing by 100

$\frac{{(x - 1)}^{2}}{25} - \frac{y^{2}}{4} = 1$ $(x-1)^2/25-y^2/4=1$

$\frac{{(x - 1)}^{2}}{5^{2}} - \frac{y^{2}}{2^{2}} = 1$ $(x-1)^2/5^2-y^2/2^2=1$
This is the standard equation of a left right hyperbola
$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x-h)^2/a^2-(y-k)^2/b^2=1$

so, the center is $(h, k) = (1, 0)$ $(h,k)=(1,0)$
the vertices ae $(h + a, k) = (1 + 5, 0) = (6, 0)$ $(h+a,k)=(1+5,0)=(6,0)$
and $(h - a, k) = (1 - 5, 0) = (- 4, 0)$ $(h-a,k)=(1-5,0)=(-4,0)$
The slopes of the asymptotes are $\pm \frac{b}{a} = \pm \frac{2}{5}$ $+-b/a=+-2/5$
And the equations of the asymptotes are $y = \pm \frac{b}{a} (x - h)$ $y=+-b/a(x-h)$
$\Rightarrow$ $=>$ $y = \pm \frac{2}{5} (x - 1)$ $y=+-2/5(x-1)$
To determine the foci, we need $c = \pm \sqrt{a^{2} + b^{2}} = \pm \sqrt{25 + 4} = \pm \sqrt{29}$ $c=+-sqrt(a^2+b^2)=+-sqrt(25+4)=+-sqrt29$
So the foci are $(h + c, k) = (1 + \sqrt{29}, 0)$ $(h+c,k)=(1+sqrt29,0)$
and $(h - c, k) = (1 - \sqrt{29}, 0)$ $(h-c,k)=(1-sqrt29,0)$

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How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola $4 x^{2} - 25 y^{2} - 8 x - 96 = 0$ $4x^2-25y^2-8x-96=0$ ?

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Explanation:

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How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola 4x2−25y2−8x−96=04x^2-25y^2-8x-96=0?

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How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola $4 x^{2} - 25 y^{2} - 8 x - 96 = 0$ $4x^2-25y^2-8x-96=0$ ?