Let e_1 and e_2 be the eccentricities of the ellipse x^225+y^2b^2=1 (b

Question

Let $e_1$ and $e_2$ be the eccentricities of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{b^2}=1$ ($b<5$) and the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{b^2}=1$ respectively, satisfying $e_1e_2=1$. 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:

Accepted Answer

$e_1^2=1-\dfrac{b^2}{25}$, $e_2^2=1+\dfrac{b^2}{16}$. $(e_1e_2)^2=1$:
$$\left(1-\frac{b^2}{25}\right)\left(1+\frac{b^2}{16}\right)=1\ \Longrightarrow\ b^2=9.$$

$$e_1=\sqrt{1-\frac9{25}}=\frac45,\qquad e_2=\sqrt{1+\frac9{16}}=\frac54.$$

$$\alpha=2(5)e_1=8,\qquad \beta=2(4)e_2=10.$$

Solutions · 1