Let and be the eccentricities of the ellipse and the hyperbola , respectively. If b

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 b < 5 and $e_{1} e_{2} = 1$ , 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 : [2025]

Smart Achievers · Accepted Answer

(3)

We have, $e_{1}^{2} = 1 - \frac{b^{2}}{25} and e_{2}^{2} = 1 + \frac{b^{2}}{16}$

$∵ e_{1} e_{2} = 1$ (Given)

$∴ e_{1}^{2} e_{2}^{2} = 1$

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

$\Rightarrow 1 + \frac{b^{2}}{16} - \frac{b^{2}}{25} - \frac{b^{4}}{400} = 1$

$\Rightarrow \frac{9 b^{2}}{400} = \frac{b^{4}}{400}$

$\Rightarrow b^{2} = 9$

We get, $\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1 and \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

Here, $e_{1} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} and e_{2} = \sqrt{1 + \frac{9}{16}} = \frac{5}{4}$

Foci : $(0, \pm 4) and (\pm 5, 0)$

The ellipse passing through all four foci is $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$

Now, the eccentricity is given by $e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$ .

Solution

1 $\frac{\sqrt{3}}{2}$

2 $\frac{4}{5}$

3 $\frac{3}{5}$

4 $\frac{\sqrt{7}}{4}$

Want Us to Email you About Special Offers & Updates?

Solution

1 32

2 45

3 35

4 74

Want Us to Email you About Special Offers & Updates?

1 $\frac{\sqrt{3}}{2}$

2 $\frac{4}{5}$

3 $\frac{3}{5}$

4 $\frac{\sqrt{7}}{4}$