# Find the equations of the hyperbola satisfying the given conditions : Foci $(\pm 5, 0)$, the transverse axis is of length $8$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{16}-\large\frac{y^2}{9}=1 & \quad (B)\;\large\frac{y^2}{16}-\large\frac{x^2}{9}=1 \\ (C)\;\large\frac{x^2}{16}+\large\frac{y^2}{9}=1 & \quad (D)\;\large\frac{x^2}{9}+\large\frac{y^2}{16}=1 \end {array}$

Toolbox:
• Standard equation of a hyperbola where transverse axis is along x - axis is $\large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2}$$=1$
• Standard equation of a hyperbola whose transverse axis is along y - axis is $\large\frac{y^2}{a^2} $$- \large\frac{x^2}{b^2}$$=1$
• $c^2=a^2+b^2$, where $c$ is the foci
• Eccentricity $e = \large\frac{c}{a}$
• Length of the latus rectum $= \large\frac{2b^2}{a}$
Step 1 :
It is given that the coordinates of the foci are $(\pm 5, 0)$ and the transverse axis is of length 8.
The foci are on the x - axis.
$\therefore$ The equation of the parabola should be of the form
$\large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2}$$=1$
Length of the transverse axis is 8.
$\Rightarrow 2a = 8 \quad \therefore a = 4$
$c^2 = a^2+b^2$
Substituting the values for c and a we get,
$25 = 16+b^2$
$\Rightarrow b^2 = 25-16 = 9$
Hence equation of the hyperbola is
$\large\frac{x^2}{16}$$-\large\frac{y^2}{9}$$=1$