Let $e_{1}$ and $e_{2}$ be the eccentricities of the ellipse,
$\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1(b<5)$ and 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: , 4
Equation of ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1$
Then, $e_{1}=\sqrt{1-\frac{b^{2}}{25}}$
The equation of hyperbola, $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$
Then, $e_{2}=\sqrt{1+\frac{b^{2}}{16}}$
$e_{1} e_{2}=1$
$\Rightarrow\left(e_{1} e_{2}\right)^{2}=1 \Rightarrow\left(1-\frac{b^{2}}{25}\right)\left(1+\frac{b^{2}}{16}\right)=1$
$\Rightarrow 1+\frac{b^{2}}{16}-\frac{b^{2}}{25}-\frac{b^{4}}{25 \times 16}=1$
$\Rightarrow \frac{9}{16 \cdot 25} b^{2}-\frac{b^{4}}{25 \cdot 16}=0 \Rightarrow b^{2}=9$
$\therefore e_{1}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$
And, $e_{2}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$
Distance between focii of ellipse
$=\alpha=2 a e_{1}=2(5)\left(e_{1}\right)=8$
Distance between focii of hyperbola
$=\beta=2 a e_{2}=2(4)\left(e_{2}\right)=10$
$\therefore(\alpha, \beta)=(8,10)$
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