Question

Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .

2x² − 3y² = 5.

Answer 1

Equation of the hyperbola: 2x² − 3y² = 5.

This can be rewritten in the following manner:

\[\frac{2 x^2}{5} - \frac{3 y^2}{5} = 1\]

\[ \Rightarrow \frac{x^2}{\frac{5}{2}} - \frac{y^2}{\frac{5}{3}} = 1\]

This is the standard equation of a hyperbola, where 

\[a^2 = \frac{5}{2} \text { and }b^2 = \frac{5}{3}\] .

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

\[ \Rightarrow \frac{5}{3} = \frac{5}{2}( e^2 - 1)\]

\[ \Rightarrow e^2 - 1 = \frac{2}{3}\]

\[ \Rightarrow e^2 = \frac{5}{3}\]

\[ \Rightarrow e = \sqrt{\frac{5}{3}}\]

Coordinates of the foci are given by  \[\left( \pm ae, 0 \right)\], i.e.

\[\left( \pm \frac{5\sqrt{6}}{6}, 0 \right)\].

Equation of the directrices: \[x = \pm \frac{a}{e}\]

\[x = \pm \frac{\sqrt{\frac{5}{2}}}{\sqrt{\frac{5}{3}}}\]

\[ \Rightarrow x = \pm \frac{\sqrt{3}}{\sqrt{2}}\]

\[ \Rightarrow \sqrt{2}x \pm \sqrt{3} = 0\]

Length of the latus rectum of the hyperbola is \[\frac{2 b^2}{a}\].

\[\Rightarrow \frac{2 \times \left( \frac{5}{3} \right)}{\sqrt{\frac{5}{2}}} = \frac{10}{3}\sqrt{\frac{2}{5}}\]