Question

Let $e_1$ and $e_1$ be the eccentricities of the ellipse, $\left(\frac{x^2}{25}\right)+\left(\frac{y^2}{b^2}\right)=1\left(b<5\right)$ and the hyperbola, $\left(\frac{x^2}{16}\right)-\left(\frac{y^2}{b^2}\right)=1$ 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 $\left(\alpha ,\beta \right)$ is equal to:

• A. $\left(8,12\right)$
• B. $\left(\frac{24}{5},10\right)$
• C. $\left(\frac{20}{3},12\right)$
• D. $\left(8,10\right)$

Answer 1

The correct option is D

$\left(8,10\right)$

Explanation for correct option:

Finding the ordered pair:

Step 1: Calculating eccentricity:

Given Data

$\frac{x^2}{25}+\frac{y^2}{b^2}=1\left(b<5\right)$ is equation of ellipse.

$\frac{x^2}{16}-\frac{y^2}{b^2}=1$ is equation of hyperbola.

$e_1{e}_2=1$

For the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1\left(b<5\right)$

Compare this equation with standard equation of ellipse $\frac{x^2}{{a_1}^2}+\frac{y^2}{b^2}=1$.

$\Rightarrow {a_1}^2=25$

Let $e_1$ is an eccentricity of ellipse.

$$\begin{gathered} e_1=\sqrt{1-\frac{b^2}{{a_1}^2}} \\ \Rightarrow e_1=\sqrt{1-\frac{b^2}{25}} \end{gathered}$$

For hyperbola $\frac{x^2}{16}-\frac{y^2}{b^2}=1$

Compare this equation with standard equation of ellipse $\frac{x^2}{{a_2}^2}-\frac{y^2}{b^2}=1$.

$\Rightarrow {a_2}^2=16$

Let $e_2$ is an eccentricity of hyperbola.

$$\begin{gathered} e_2=\sqrt{1+\frac{b^2}{{a_2}^2}} \\ \Rightarrow e_2=\sqrt{1+\frac{b^2}{16}} \end{gathered}$$

Step 2: Determining the value of $b$:

Since $e_1{e}_2=1$,

$$\begin{gathered} e_1{e}_2=1 \\ \Rightarrow \left(\sqrt{1-\frac{b^2}{25}}\right)\left(\sqrt{1+\frac{b^2}{16}}\right)=1 \\ \Rightarrow \left(1-\frac{b^2}{25}\right)\left(1+\frac{b^2}{16}\right)=1^2 \\ \Rightarrow 1+\frac{b^2}{16}-\frac{b^2}{25}-\frac{b^4}{25\times 16}=1 \\ \Rightarrow 1+\frac{9b^2-b^4}{16\times 25}=1 \\ \Rightarrow 9b^2-b^4=0 \\ \Rightarrow 9-b^2=0 \\ \Rightarrow b^2=9 \end{gathered}$$

Step 3: Calculating eccentricity:

Substitute $b^2=9$ in $e_1=\sqrt{1-\frac{b^2}{25}}$

$$\begin{gathered} \Rightarrow e_1=\sqrt{1-\frac{b^2}{25}} \\ \Rightarrow e_1=\sqrt{1-\frac{16}{25}} \\ \Rightarrow e_1=\sqrt{\frac{9}{25}} \\ \Rightarrow e_1=\frac{3}{5} \end{gathered}$$

Substitute $b^2=9$ in $e_2=\sqrt{1+\frac{b^2}{16}}$

$$\begin{gathered} \Rightarrow e_2=\sqrt{1+\frac{b^2}{16}} \\ \Rightarrow e_2=\sqrt{1+\frac{9}{16}} \\ \Rightarrow e_2=\sqrt{\frac{25}{16}} \\ \Rightarrow e_2=\frac{5}{4} \end{gathered}$$

Step 4: Determining the ordered pair:

The distance between the foci is $2ae$

The distance for the ellipse

$$\begin{gathered} \alpha =2\times a_1\times e_1 \\ \Rightarrow \alpha =\left(2\times 5\times \frac{4}{5}\right) \\ \Rightarrow \alpha =8 \end{gathered}$$

The distance for the hyperbola

$$\begin{gathered} \beta =2\times a_2\times e_2 \\ \Rightarrow \beta =\left(2\times 4\times \frac{5}{4}\right) \\ \Rightarrow \beta =10 \end{gathered}$$

Hence, $\left(\alpha ,\beta \right)=\left(8,10\right)$

Therefore, the correct answer is option (D).