Given equation of ellipse is 16x2+b2y2=1 eccentricity =e=1−16b2 foci: ±ae=±41−16b2 Equation of hyperbola is 144x2−81y2=251 ⇒25144x2−2581y2=1 eccentricity =e=1+2581×14425=1+14481=144225=1215 foci: ±ae=±512×1215=±3 Since, foci of ellipse and hyperbola coincide & \therefore \pm 4 \sqrt{1-\frac{b^2}{16}}=\pm 3 \Rightarrow b^2=7 \ $$