Ellipse E: x²/36 + y²/16 = 1. A hyperbola confocal with ellipse E and eccentricity of hyperbola is equal to 5. The length of latus rectum of the hyperbola is, if the principal axis of hyperbola is the x-axis?

To find the length of the latus rectum of the hyperbola confocal with the given ellipse, we proceed step-by-step:

Step 1: Find the foci of the given ellipse
The equation of the ellipse is given by:
$\frac{x^2}{36}+\frac{y^2}{16}=1$
Comparing this with the standard equation of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we have:
$a^2=36\Rightarrow a=6$
$b^2=16\Rightarrow b=4$

Let $e_e$ be the eccentricity of the ellipse. The relationship between the semi-axes and eccentricity is:
$b^2=a^2(1-e_e^2)$
Substituting the values:
$16=36(1-e_e^2)$
$1-e_e^2=\frac{16}{36}=\frac{4}{9}$
$e_e^2=1-\frac{4}{9}=\frac{5}{9}$
$e_e=\frac{\sqrt{5}}{3}$

The coordinates of the foci of the ellipse are given by $(\pm a{e}_e,0)$:
$a{e}_e=6·\frac{\sqrt{5}}{3}=2\sqrt{5}$
Thus, the foci of the ellipse are at $(\pm 2\sqrt{5},0)$.

Step 2: Determine the equation parameters of the hyperbola
Since the hyperbola is confocal with the ellipse and its principal axis is the x-axis, its foci are also $(\pm 2\sqrt{5},0)$.
Let the equation of the hyperbola be:
$\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$

Let $e_h$ be the eccentricity of the hyperbola. We are given that $e_h=5$.
For a confocal hyperbola, the focal distance is the same:
$A{e}_h=2\sqrt{5}$
Substituting $e_h=5$:
$5A=2\sqrt{5}\Rightarrow A=\frac{2\sqrt{5}}{5}=\frac{2}{\sqrt{5}}$
Thus, the square of the semi-transverse axis is:
$A^2=\frac{4}{5}$

For a hyperbola, the relation between the semi-axes and eccentricity is:
$B^2=A^2(e_h^2-1)$
Substituting the values of $A^2$ and $e_h$:
$B^2=\frac{4}{5}(5^2-1)$
$B^2=\frac{4}{5}\left(24\right)$
$B^2=\frac{96}{5}$

Step 3: Calculate the length of the latus rectum of the hyperbola
The length of the latus rectum of the hyperbola is given by the formula:
$\text{L.R.}=\frac{2B^2}{A}$
Substituting the values of $B^2$ and $A$:
$\text{L.R.}=\frac{2·\left(\frac{96}{5}\right)}{\frac{2}{\sqrt{5}}}$
$\text{L.R.}=2·\frac{96}{5}·\frac{\sqrt{5}}{2}$
$\text{L.R.}=\frac{96\sqrt{5}}{5}$
Simplifying the expression by dividing the numerator and denominator by $\sqrt{5}$:
$\text{L.R.}=\frac{96}{\sqrt{5}}$

Therefore, the length of the latus rectum of the hyperbola is $\frac{96}{\sqrt{5}}$.