The foci of the ellipse $$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$$   and the hyperbola $$\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$$    coincide. Then the value of $${{b^2}}$$  is-

Solution :
$$\eqalign{ & \frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}} \cr & a = \sqrt {\frac{{144}}{{25}}} ,\,b = \sqrt {\frac{{81}}{{25}}} ,\,e = \sqrt {1 + \frac{{81}}{{144}}} = \frac{{15}}{{12}} = \frac{5}{4} \cr} $$
$$\therefore $$ Foci $$ = \left( { \pm 3,\,0} \right)$$
$$\therefore $$ foci of ellipse $$=$$ foci of hyperbola
$$\therefore $$ for ellipse $$ae = 3$$   but $$a=4,$$
$$\therefore e = \frac{3}{4}$$
Then $${b^2} = {a^2}{\left( {1 - e} \right)^2}\,\,\,\,\,\,\, \Rightarrow {b^2} = 16\left( {1 - \frac{9}{{16}}} \right) = 7$$