(a) Given hyperbola is 

9x² - 16y² - 144

⇒ \(\frac{x^2}{16}-\frac{y^2}9=1\)

By comparing with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

We get a² = 16, b² = 9

⇒ a = 4, b = 3

\(\therefore\) e = \(\sqrt{1+\frac{b^2}{a^2}}\) = \(\sqrt{1+\frac{3^2}{4^2}}\)

\(=\frac{4^2+3^2}{4^2}\) = \(\frac{5^2}{4^2}=\frac54\) 

\(\therefore\) coordinates of fozi are

(+ae, 0) = (\(\pm\)4 x 5/4,0) = (\(\pm5,0\))

Hence, (-5, 0) and (5, 0) are coordinates of foci or given hyperbola.

And latus rectum of hyperbola = \(\frac{2b^2}a=\frac{2\times9}4=\frac92\) unit

(b) Given ellipse is 3x² + 2y² = 18 

⇒ \(\frac{x^2}6+\frac{y^2}9=1\)

\(\therefore\) a² = 6 and b² = 9

⇒ a = √6 and b = 3

\(\because\) b > a

\(\therefore\) Focus of ellipse lies on y-axis.

Now, e = \(\sqrt{1-\frac{q^2}{b^2}}=\sqrt{1-\frac69}\) = \(\sqrt{\frac{9-6}9}\) = \(\frac{\sqrt3}3\) = \(\frac1{\sqrt3}\) 

\(\therefore\) Co-ordinates of foci are

(0, \(\pm\)be) = (0, \(\pm\)9 x \(\frac1{\sqrt3}\)) = (0, \(\pm\)\(\frac{3}{\sqrt3}\)) (0, \(\pm\sqrt3\))

Hence, (0, \(\sqrt3\)) and (0, \(-\sqrt3\)) are co-ordinates of foci or given ellipse.

And latus rectum of ellipse = \(\frac{2b^2}a=\frac{2\times3}{\sqrt6}\) = \(\frac{6}{\sqrt6}\) = \(\sqrt6\) unit