Hyperbola

Given, ellipse is $\;\frac{x^{2}}{25}+\;\frac{y^{2}}{16}=1$ or $\;\; \;\frac{x^{2}}{(5)^{2}}+\;\frac{y^{2}}{(4)^{2}}=1$ Compare it with standard equation $\;\frac{x^{2}}{a^{2}}+\;\frac{y^{2}}{b^{2}}=1$, we get $a=5, b=4$ Now, focus of ellipse $=(\pm C, 0)$ where $C=\sqrt{a^{2}-b^{2}}$ Put the values of $a$ and $b$, we get $C=\sqrt{5^{2}-4^{2}}=\sqrt{25-16}=\sqrt{9}$ $\therefore$ Focus $=(\pm 3,0)$ According to question, hyperbola passes through the focus of ellipse. Let equation of hyperbola be $\;\frac{x^{2}}{a^{2}}-\;\frac{y^{2}}{b^{2}}=1$ Since, it passes through $(\pm 3,0)$, we get $\;\frac{(\pm 3)^{2}}{a^{2}}-\;\frac{0}{b^{2}}=1$, gives $a=\pm 3$ or $a^{2}=9$ Also, given that product of eccentricities is $1 .$ Now, (Eccentricity of ellipse ) (Eccentricity of hyperbola) $=1$ $\Rightarrow \;\; \left(\sqrt{1-\;\frac{16}{25}}\right)\left(\sqrt{1+\;\frac{b^{2}}{9}}\right)=1$ (using formula of eccentricity of ellipse and hyperbola) $\Rightarrow \;\; \left(\sqrt{\;\frac{9}{25}}\right)\left(\sqrt{1+\;\frac{b^{2}}{9}}\right)=1$ Squaring on both sides, $1+\;\frac{b^{2}}{9}=\;\frac{25}{9}$ $b^{2}=16$ Thus, equation of hyperbola is $\;\frac{x^{2}}{9}-\;\frac{y^{2}}{16}=1$.