Let e1 and e2 be the eccentricities of the ellipse x^2/b^2 + y^2/25 = 1 and the hyperbola x^2/16 - y^2/b^2 = 1, respectively.

Question

Let \(e_{1}\) and \(e_{2}\) be the eccentricities of the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{25}=1 \) and the hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1,\) respectively. If \(\mathrm{b}<5\) and \(\mathrm{e}_{1} \mathrm{e}_{2}=1,\) 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 :

(1) \(\frac{4}{5}\)

(2) \(\frac{3}{5}\)

(3) \(\frac{\sqrt{7}}{4}\)

(4) \(\frac{\sqrt{3}}{2}\)

Answer 1

Correct option is: (2) \(\frac{3}{5}\)   

\(\mathrm{e}_{1}^{2}=1-\frac{\mathrm{b}^{2}}{25} \quad \mathrm{e}_{2}^{2}=1-\frac{\mathrm{b}^{2}}{16}\)

\(\therefore \mathrm{e}_{1}^{2} \mathrm{e}_{2}^{2}=1\)

\(\left(1-\frac{\mathrm{b}^{2}}{25}\right)\left(1+\frac{\mathrm{b}^{2}}{16}\right)=1\)

\(\Rightarrow 2+\frac{\mathrm{b}^{2}}{16}-\frac{\mathrm{b}^{2}}{25}-\frac{\mathrm{b}^{2}}{400}=1\)

\(\Rightarrow \frac{9 \mathrm{~b}^{2}}{400}=\frac{\mathrm{b}^{4}}{400}\)

\(b^{2}=9\)

\(\frac{x^{2}}{9}+\frac{y^{2}}{25}=1 \quad \frac{x^{2}}{16}-\frac{y^{2}}{9}=0\)

\(e_{1} \sqrt{1-\frac{9}{25}} \quad e_{2}=\frac{5}{4}\)

\(e_{1}=\frac{4}{5}\)

Focii : - \((0, \pm 4) \quad( \pm 5,0)\)

ellipse passing through all four foci

\(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)

\(e=\sqrt{1-\frac{16}{25}}=\frac{3}{5}\)