Question

$\text{Find the axes,eccentricity, latus-rectum and the co-ordinates of the foci of the hyperbola.}25x^2-36y^2=225$

Answer 1

$Wehave,25x^2-36y^2=225\Rightarrow \frac{25x^2}{225}-\frac{36y^2}{225}=1\Rightarrow \frac{x^2}{9}-\frac{4y^2}{25}=1\Rightarrow \frac{x^2}{9}-\frac{y^2}{\frac{25}{4}}=1\text{This is of the form}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,wherea=3andb=\frac{5}{2}\text{Length of the transverse aixis : The length of the transverse axis =2a}=2\times 3=6\text{Length of the conjugate axis : The length of the conjugate axis is}2b=2\times \frac{5}{2}=5\text{Eccentricity : The eccentricity e is given by}e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{\frac{25}{4}}{9}}=\sqrt{1+\frac{25}{36}}=\sqrt{\frac{61}{36}}=\frac{\sqrt{6}1}{6}\text{Length of Latus-rectum}=\frac{2b^2}{a}=\frac{25}{6}Foci(\pm \frac{\sqrt{6}1}{2},0)$