The vertices of a hyperbola are at (0, 0) and | vedclass

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(B) The vertices are $(0, 0)$ and $(10, 0)$. The center of the hyperbola is the midpoint of the vertices: $(\frac{0+10}{2}, 0) = (5, 0)$.The distance

Vedclass · Accepted Answer

$\frac{(x - 5)^2}{25} - \frac{y^2}{144} = 1$

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(B) The vertices are $(0, 0)$ and $(10, 0)$. The center of the hyperbola is the midpoint of the vertices: $(\frac{0+10}{2}, 0) = (5, 0)$.The distance between the vertices is $2a = 10$,so $a = 5$.The distance from the center $(5, 0)$ to the focus $(18, 0)$ is $ae = 18 - 5 = 13$.Since $a = 5$,we have $5e = 13$,so $e = \frac{13}{5}$.Using the relation $b^2 = a^2(e^2 - 1)$,we get $b^2 = 25(\frac{169}{25} - 1) = 169 - 25 = 144$,so $b = 12$.The equation of the hyperbola with center $(h, k) = (5, 0)$ is $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$.Substituting the values,we get $\frac{(x - 5)^2}{25} - \frac{y^2}{144} = 1$.

The vertices of a hyperbola are at $(0, 0)$ and $(10, 0)$ and one of its foci is at $(18, 0)$. The equation of the hyperbola is:

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