Let $\mathrm{e}_{1}$ and $\mathrm{e}_{2}$ be the eccentricities of the ellipse,
$\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1(b<5) \quad$ and $\quad$ the hyperbola,
$\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1$ respectively satisfying $e_{1} e_{2}=1$. 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 :
Correct Option: 1
For ellipse $\frac{\mathrm{x}^{2}}{25}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ $(b<5)$
Let $\mathrm{e}_{1}$ is eccentricity of ellipse
$\therefore \quad b^{2}=25\left(1-e_{1}^{2}\right) \ldots \ldots .$(i)
Again for hyperbola
$\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1$
Let $e_{2}$ is eccentricity of hyperbola.
$\therefore \quad \mathrm{b}^{2}=16\left(\mathrm{e}_{2}{ }^{2}-1\right)$......(2)
by (1) \& (2)
$25\left(1-\mathrm{e}_{1}^{2}\right)=16\left(\mathrm{e}_{2}^{2}-1\right)$
Now $e_{1} \cdot e_{2}=1$ (given)
$\therefore \quad 25\left(1-e_{1}^{2}\right)=16\left(\frac{1-e_{1}^{2}}{e_{1}^{2}}\right)$
or $\quad e_{1}=\frac{4}{5}$ $\therefore \mathrm{e}_{2}=\frac{5}{4}$
Now distance between foci is $2 \mathrm{ae}$
$\therefore$ distance for ellipse $=2 \times 5 \times \frac{4}{5}=8=\alpha$
distance for hyperbola $=2 \times 4 \times \frac{5}{4}=10=\beta$
$\therefore(\alpha, \beta) \equiv(8,10)$
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