Let e1 and e2 be the eccentricities of the ellipse (x^2)/(b^2) + (y^2)/(25) = 1 and the hy…

Question

Let e1 and e2 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 e1e2 = 1, then the eccentricity of the ellipse having its axes al...

Accepted Answer

Let's find the eccentricities of the given ellipse and hyperbola, and then determine the eccentricity of an ellipse that passes through all four foci.

Find ${e}_{1}$ for the Ellips: 1: Find ${e}_{1}$ for the Ellipse

The equation of the ellipse is:

$\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{25}=1$

The eccentricity ${e}_{1}$ is given by:

${e}_{2}^{1}=1-\frac{{b}^{2}}{25}$

Find ${e}_{2}$ for the Hyperbol: 2: Find ${e}_{2}$ for the Hyperbola

The equation of the hyperbola is:

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

The eccentricity ${e}_{2}$ is given by:

${e}_{2}^{2}=1+\frac{{b}^{2}}{16}$

Using the Product ${e}_{1}{e}_{2}=1: 3: Using the Product ${e}\_{1}{e}\_{2}=1$

Given:

${e}\_{1}{e}\_{2}=1$

Thus:

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

Expanding gives:

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

Simplifying:

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

Thus:

${b}\^{2}=9$

Determine Eccentricities ${e}\_{1}$ and ${e}\_{2}: 4: Determine Eccentricities ${e}_{1}$ and ${e}_{2}$

Substitute ${b}^{2}=9$:

For the ellipse:

${e}_{2}^{1}=1-\frac{9}{25}=\frac{16}{25}$

${e}_{1}=\frac{4}{5}$

For the hyperbola:

${e}_{2}=\frac{5}{4}$

Find the Eccentricity of the New Ellips: 5: Find the Eccentricity of the New Ellipse

The new ellipse's equation is:

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

The eccentricity $e$ is:

$e=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$

Thus, the eccentricity of the ellipse that passes through all four foci is $\frac{3}{5}$.

Solutions · 1