Find the value

Given Equation: $24 x^{2}-25 y^{2}=600 \Rightarrow$

$\frac{x^{2}}{25}-\frac{y^{2}}{24}=1$

Comparing with the equation of hyperbola

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ we get,

$a=5$ and $b=\sqrt{24}=2 \sqrt{6}$

(i) Length of Transverse axis $=2 a=10$ units.

Length of Conjugate axis $=2 b=4 \sqrt{6}$ units.

(ii) Coordinates of the vertices $=(\pm a, 0)=(\pm 5,0)$

(iv) Here, eccentricity, $e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{24}{25}}=\sqrt{\frac{49}{25}}=\frac{7}{5}$

(iii) Coordinates of the foci $=(\pm a e, 0)=(\pm 7,0)$

(v) Length of the rectum $=\frac{2 b^{2}}{a}=\frac{48}{5}=9.6$ units.