Let e_{1} and e_{2} be the eccentricities of | vedclass

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Question

For ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1 \quad(b<5)$

Let e $_{1}$ is eccentricity of ellipse

$	herefore \quad b^{2}=25\left(1-e_{

Vedclass · Accepted Answer

$(8,10)$

Vedclass · Answer

For ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1 \quad(b<5)$

Let e $_{1}$ is eccentricity of ellipse

$	herefore \quad b^{2}=25\left(1-e_{1}^{2}ight) \ldots \ldots .(1)$

Again for hyperbola

$\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1$

Let $e _{2}$ is eccentricity of hyperbola.

$	herefore \quad b^{2}=16\left(e_{2}^{2}-1ight) \quad \ldots \ldots$

by (1)$\&(2)$

$25\left(1- e _{1}^{2}ight)=16\left( e _{2}^{2}-1ight)$

Now $e _{1} e _{2}=1 \quad$ (given)

$	herefore \quad 25\left(1- e _{1}^{2}ight)=16\left(\frac{1- e _{1}^{2}}{ e _{1}^{2}}ight)$

or $\quad e _{1}=\frac{4}{5} \quad 	herefore e _{2}=\frac{5}{4}$

Now distance between foci is $2 ae$

$	herefore$ distance for ellipse $=2 	imes 5 	imes \frac{4}{5}=8=\alpha$

distance for hyperbola $=2 	imes 4 	imes \frac{5}{4}=10=\beta$

$	herefore(\alpha, \beta) \equiv(8,10)$

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