# Let the function

Question:

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 :

1. $(8,10)$

2. $(8,12)$

3. $\left(\frac{20}{3}, 12\right)$

4. $\left(\frac{24}{5}, 10\right)$

Correct Option: 1

Solution:

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 $=2 \times 4 \times \frac{5}{4}=10=\beta$

$\therefore(\alpha, \beta) \equiv(8,10)$