Find the eccentricity, center, foci and vertices of the hyperbola $25x^2 - 16y^2 = 400$ and sketch it.
Equation of given hyperbola is: $\;$ $25x^2 - 16y^2 = 400$
i.e. $\;$ $\dfrac{x^2}{400 / 25} - \dfrac{y^2}{400 / 16} = 1$
i.e. $\;$ $\dfrac{x^2}{16} - \dfrac{y^2}{25} = 1$ $\;\;\; \cdots \; (1)$
The transverse axis is along the X axis.
Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives
$a^2 = 16 \implies a = 4$; $\;$ and $\;$ $b^2 = 25 \implies b = 5$
Now, eccentricity $= e = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{25}{16}} = \dfrac{\sqrt{41}}{4}$
Center $= C \left(0, 0\right)$
Foci [$F_1$ and $F_2$] $= \left(\pm ae, 0\right) = \left(\pm 4 \times \dfrac{\sqrt{41}}{4}, 0\right) = \left(\pm \sqrt{41}, 0\right)$
Vertices [$A_1$ and $A_2$] $= \left(\pm a, 0\right) = \left(\pm 4, 0\right)$
