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: \[25 x^2 - 36 y^2 = 225\]

This equation can be rewritten in the following way:

\[\frac{25 x^2}{225} - \frac{36 y^2}{225} = 1\]

\[ \Rightarrow \frac{x^2}{9} - \frac{y^2}{\frac{225}{36}} = 1\]

This is the standard equation of the hyperbola, where

\[a^2  = 9  \text { and }   b^2  = \frac{225}{36}\]

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 \[b^2 = a^2 ( e^2 - 1)\].

\[\Rightarrow \frac{225}{36} = 9\left( e^2 - 1 \right)\]

\[ \Rightarrow e^2 - 1 = \frac{25}{36}\]

\[ \Rightarrow e^2 = \frac{61}{36}\]

\[ \Rightarrow e = \frac{\sqrt{61}}{6}\]

Length of the latus rectum = \[\frac{2 b^2}{a} = \frac{2 \times \left( \frac{225}{36} \right)}{3} = \frac{25}{6}\]

The coordinates of the foci are given by \[\left( \pm ae, 0 \right)\].

\[\Rightarrow \left( \pm \frac{\sqrt{61}}{2}, 0 \right)\]