If the foci of the ellipse $$\frac{x^2}{16} + \frac{y^2}{b^2} = 1$$ coincide with the foci of the hyperbola $$\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$$, then $$b^2$$ is equal to

The given ellipse is $$\frac{x^2}{16}+\frac{y^2}{b^2}=1$$. Its centre is at the origin and its axes are along the coordinate axes.

The given hyperbola is $$\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$$. Multiply both sides by 25 to put it in standard form:

$$25\left(\frac{x^2}{144}-\frac{y^2}{81}\right)=1 \;\Longrightarrow\; \frac{x^2}{\tfrac{144}{25}}-\frac{y^2}{\tfrac{81}{25}}=1.$$

Hence for the hyperbola
$$a_h^2=\frac{144}{25},\qquad b_h^2=\frac{81}{25}.$$

For a hyperbola of the form $$\frac{x^2}{a_h^2}-\frac{y^2}{b_h^2}=1,$$ the focal distance satisfies $$c_h^2=a_h^2+b_h^2.$$

Therefore $$c_h^2=\frac{144}{25}+\frac{81}{25}=\frac{225}{25}=9 \;\Longrightarrow\; c_h=3.$$

Thus the foci of the hyperbola are $$(\pm3,0).$$ By the condition in the question, the ellipse must have these same foci.

For an ellipse of the form $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ with its major axis along the $$x$$-axis, the focal distance satisfies $$c^2=a^2-b^2.$$

In the ellipse under consideration, the denominator of $$x^2$$ is 16, so $$a^2=16.$$ (The other denominator $$b^2$$ is unknown, and we will soon confirm that $$b^2\lt16$$, ensuring the major axis is indeed along $$x$$.)

Since the ellipse shares the same foci as the hyperbola, we must have $$c=3.$$ Hence

$$c^2=a^2-b^2 \;\Longrightarrow\; 9=16-b^2 \;\Longrightarrow\; b^2=16-9=7.$$

The result $$b^2=7$$ is less than 16, so the major axis of the ellipse is along the $$x$$-axis, consistent with our assumption.

Therefore $$b^2=7.$$

Option C which is: $$7$$