Find the equations of the hyperbola satisfying the given conditions : Foci $(\pm 3\sqrt 5 , 0)$, the latus rectum is of length $8$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{25}+\large\frac{y^2}{20}=1 & \quad (B)\;\large\frac{x^2}{20}+\large\frac{y^2}{25}=1 \\ (C)\;\large\frac{x^2}{25}-\large\frac{y^2}{20}=1 & \quad (D)\;\large\frac{y^2}{25}-\large\frac{x^2}{20}=1 \end {array}$

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 :
The coordinates of the foci given are $( \pm 3 \sqrt 5 , 0)$ and
The length of the latus rectum is 8
Since the foci is on the x - axis.
$\therefore$ Equation of the hyperbola is
$\large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2}$$=1$
Length of the latus rectum is $\large\frac{2b^2}{a}$
(i.e) $\large\frac{2 \times b^2}{a}$$=8 \Rightarrow b^2=4a Step 2 : We know that c^2=a^2+b^2 (i.e) (3\sqrt 5 )^2=a^2+4a \Rightarrow 45 = a^2+4a or a^2+4a-45=0 On factorizing we get, a^2+9a-5a-45=0 a(a+9)-5(a+9)=0 \Rightarrow (a-5)(a+9)=0 \therefore a = 5 or a =-9 But a=-9 is not admissable. \therefore a = 5 \Rightarrow b^2=4a = 4 \times 5 =20 Hence the equation of the hyperbola is \large\frac{x^2}{25}$$-\large\frac{y^2}{20}=1$