The ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ and the hyperbola $\frac{x^{2}}{25}-\frac{y^{2}}{16}=1$ have in common

Equation of ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1, a>b$ and equation of hyperbola is $\frac{x^{2}}{25}-\frac{y^{2}}{16}=1, a>b$ Let $e$ and $e^{\prime}$ be the eccentricities of the ellipse and hyperbola. So, $e=\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\sqrt{\frac{25-16}{25}}=\frac{3}{5}$ and $\quad e^{\prime}=\sqrt{\frac{a^{2}+b^{2}}{a^{2}}}=\sqrt{\frac{25+16}{25}}=\frac{\sqrt{41}}{5}$ (i) Centre of ellipse $(0,0)$ and centre of hyperbola is $(0,0)$ (ii) Foci of ellipse are $(\pm a e, 0)$ or $(\pm 3,0)$. Foci of hyperbola are $\left(\pm a e^{\prime}, o\right)$ or $(\pm \sqrt{41}, 0)$. (iii) Direction of ellipse are $x=\pm \frac{a}{e} \Rightarrow x=\pm \frac{25}{3}$ and directrices of hyperbola are $x=\pm \frac{a}{e}$ $\Rightarrow \quad x=\pm \frac{25}{\sqrt{41}}$ (iv) Vertices of ellipse are $(\pm a, 0)$ or $(\pm 5,0)$. Vertices of hyperbola are $(\pm a, 0)$ or $(\pm 5,0)$. From the above discussions, their are common is centre and vertices.