Let e₁ and e₂ be the eccentricities of the ellipse,
${{{x^2}} \over {25}} + {{{y^2}} \over {{b^2}}} = 1$(b < 5) and the hyperbola,
${{{x^2}} \over {16}} - {{{y^2}} \over {{b^2}}} = 1$ respectively satisfying e₁e₂ = 1. If $\alpha$
and $\beta$ are the distances between the foci of the
ellipse and the foci of the hyperbola
respectively, then the ordered pair ($\alpha$, $\beta$) is equal to :

For ellipse ${{{x^2}} \over {25}} + {{{y^2}} \over {{b^2}}} = 1\,\,\,(b &lt; 5)$<br><br>Let e₁ is eccentricity of ellipse <br><br>$\therefore$ b² = 25 (1 $-$ ${e_1}^2$) ........(1)<br><br>Again for hyperbola<br><br>${{{x^2}} \over {16}} - {{{y^2}} \over {{b^2}}} = 1$<br><br>Let e₂ is eccentricity of hyperbola.<br><br>$\therefore$ ${b^2} = 16({e_1}^2 - 1)$ ......(2)<br><br>by (1) &amp; (2)<br><br>$25(1 - {e_1}^2)\, = \,16({e_1}^2 - 1)$<br><br>Now e₁ . e₂ = 1 (given)<br><br>$\therefore$ $25(1 - {e_1}^2)\, = \,16\left( {{{1 - {e_1}^2} \over {{e_1}^2}}} \right)$<br><br>or e₁ = ${4 \over 5}$ $\therefore$ e₂ = ${5 \over 4}$<br><br>Now distance between foci is 2ae<br><br>$\therefore$ Distance for ellipse = $2 \times 5 \times {4 \over 5} = 8 = \alpha$<br><br>Distance for hyperbola = $2 \times 4 \times {5 \over 4} = 10 = \beta$<br><br>$\therefore$ ($\alpha$, $\beta$) $\equiv$ (8, 10)