Question

Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x² − 36y² = 225.

Answer 1

Equation of the hyperbola:
$25x^2-36y^2=225$
This equation can be rewritten in the following way:
$$\begin{gathered} \frac{25x^2}{225}-\frac{36y^2}{225}=1 \\ \Rightarrow \frac{x^2}{9}-\frac{y^2}{{\displaystyle \raisebox{1ex}{$225$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}}=1 \end{gathered}$$
This is the standard equation of the hyperbola, where $a^2=9\mathrm{and}{b}^2=\frac{225}{36}$.
Length of the transverse = $2a=2\times 3=6$
Length of the conjugate axis = $2b=2\times \frac{15}{6}=5$

Eccentricity of the hyperbola is calculated using $b^2=a^2(e^2-1)$.

$$\begin{gathered} \Rightarrow \frac{225}{36}=9\left(e^2-1\right) \\ \Rightarrow e^2-1=\frac{25}{36} \\ \Rightarrow e^2=\frac{61}{36} \\ \Rightarrow e=\frac{\sqrt{61}}{6} \end{gathered}$$
Length of the latus rectum =$\frac{2b^2}{a}=\frac{2\times \left({\displaystyle \raisebox{1ex}{$225$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}\right)}{3}=\frac{25}{6}$
The coordinates of the foci are given by $\left(\pm ae,0\right)$.
$\Rightarrow \left(\pm \frac{\sqrt{61}}{2},0\right)$