If the focii of the ellipse (x^2)/(25)+(y^2)/(16)=1 and the...

Given, equation of ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=$1 and equation of hyperbola is $\frac{x^2}{4}-\frac{y^2}{b^2}=$1 eccentricity of ellipse $$\begin{aligned}& b²=a²\left(1-e²\right) \\ & \Rightarrow \quad 16=25\left(1-e²\right) \\ & \Rightarrow \quad e²=1-\frac{16}{25}=\frac{9}{25}\\ & \Rightarrow \quad e= \pm \frac{3}{5}\\ & \end{aligned}$$Focii of the ellipse $=( \pm a e, 0)=( \pm 3,0)$ which coincide with focii of the hyperbola. Let $e_1$ be the eccentricity of the hyperbola. $$\begin{array}{ll} \therefore& \pma e_1= \pm3 \\ \Rightarrow & e_1=\frac{3}{2} \end{array}$$Now, $\quad b^2=a^2$\left(e_1^2-1\right)$\begin{array}{cc} \Rightarrow & b^2=4\left(\frac{9}{4}-1\right)=4 \times \frac{5}{4}\\ \Rightarrow & b²=5 \end{array}$$