Let e_1 and e_2 be the eccentricities of the ellipse x^2b^2+y^225=1 and the hyperbola x^21…

Question

Let $e_1$ and $e_2$ be the eccentricities of the ellipse $\dfrac{x^2}{b^2}+\dfrac{y^2}{25}=1$ and the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{b^2}=1$, respectively. If $b<5$ and $e_1e_2=1$, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:

Accepted Answer

$$e_1^2=1-\frac{b^2}{25},\qquad e_2^2=1+\frac{b^2}{16}.$$

$e_1e_2=1\Rightarrow e_1^2e_2^2=1$:

$$\left(1-\frac{b^2}{25}\right)\left(1+\frac{b^2}{16}\right)=1 \;\Rightarrow\; \frac{9b^2}{400}=\frac{b^4}{400} \;\Rightarrow\; b^2=9.$$

$$e_1^2=1-\frac{9}{25}=\frac{16}{25}\Rightarrow e_1=\frac45,\qquad e_2=\frac54.$$

With $b^2=9$, the ellipse's foci are $(\pm be_1,0)=\left(\pm\dfrac{12}{5},0\right)$ and the hyperbola's foci are $(\pm4e_2,0)=(\pm5,0)$. The ellipse through all four foci (largest semi-axis $5$) is $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$, with

$$e=\sqrt{1-\frac{16}{25}}=\frac35.$$

Solutions · 1