In the hyperbola $4{{x}^{2}}-9{{y}^{2}}=36$, find the axes, the coordinates of the foci, the eccentricity, and the latus rectum.

Hint: Compare the given equation of hyperbola with the general equation $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$. Get the values of a and b and find the quantities mentioned.

Complete step-by-step solution:
We know the standard equation of hyperbola
$\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$……………….. (1)
From the question, the equation of hyperbola
$4{{x}^{2}}-9{{y}^{2}}=36$………………… (2)
Divide the entire equation (2) by 36
$\begin{align}
  & \dfrac{4{{x}^{2}}-9{{y}^{2}}}{36}=\dfrac{36}{36} \\
 & \Rightarrow \dfrac{4{{x}^{2}}}{36}-\dfrac{9{{y}^{2}}}{36}=1 \\
 & \Rightarrow \dfrac{{{x}^{2}}}{9}-\dfrac{{{y}^{2}}}{4}=1 \\
 & \Rightarrow \dfrac{{{x}^{2}}}{{{3}^{2}}}-\dfrac{{{y}^{2}}}{{{2}^{2}}}=1...............\left( 3 \right) \\
\end{align}$
Now compare equation (1) with equation (3).
We get a = 3 and b = 2.
The transverse axis is a line segment that passes through the centre of the hyperbola and has vertices as its end points. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co – vertices as its endpoints.
$\therefore $ Transverse axis = 2a
Conjugate axis = 2b
$\therefore $ The coordinates of the vertices are $\left( \pm \ ae, \, 0 \right)$ and co – vertices are $\left( 0,\pm \ b \right)$
$\therefore $ Transverse axis $=2a=2\times 3=6$
Conjugate axis =$ 2b = 2 \times 2 = 4$
Eccentricity of hyperbola is given by:
$\begin{align}
  & e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}},\ where\ a>b \\
&\Rightarrow e=\sqrt{1+\dfrac{{{2}^{2}}}{{{3}^{2}}}}=\sqrt{1+\dfrac{4}{9}}=\sqrt{\dfrac{9+4}{9}}=\sqrt{\dfrac{13}{9}} \\
 & \therefore e=\sqrt{\dfrac{13}{9}} \\
\end{align}$
We know the coordinates of foci as
$\begin{align}
  & S=\left( ae,0 \right)\ and\ S'=\left( -ae,0 \right) \\
 & \therefore S=\left( 3\sqrt{\dfrac{13}{9}},0 \right)\ and\ S'=\left( -3\sqrt{\dfrac{13}{9}},0 \right) \\
 & \Rightarrow \left( 3\sqrt{\dfrac{13}{9}},0 \right)=S \\
 & and\ S'=\left( -3\sqrt{\dfrac{13}{9}},0 \right) \\
 & \therefore S=\left( \sqrt{13},0 \right)\ and \ S'=\left( -\sqrt{13},0 \right) \\
\end{align}$
$\therefore \left( \pm \sqrt{13},0 \right)$ are the coordinates of foci.
The length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times {{2}^{2}}}{3}=\dfrac{2\times 4}{3}=\dfrac{8}{3}$

Note: As you are asked to find the foci and other factors of the hyperbola, it is important that you remember the basic formulas of the hyperbola. Similarly, learn the basic properties of parabolas as well. In this particular question, for the case, take the equation as, $a>b,e=\sqrt{1+\dfrac{{{a}^{2}}}{{{b}^{2}}}}$
Similarly in case of: $a