Question

Let $\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1$, $\alpha > \beta$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{3}}$. If $e_H$ denotes the eccentricity of $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$, then $e_H^2$ equals

• A. $\frac{11}{3}$
• B. $\frac{5}{3}$
• C. $\frac{12}{7}$
• D. $\frac{25}{3}$

Answer 1

Answer: (B)

$\frac{5}{3}$

Answer 2

Explanation:

1. Eccentricity of the ellipse: For an ellipse $\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1$ with $\alpha > \beta$, the eccentricity $e$ is given by the relation $e^2 = 1 - \frac{\beta^2}{\alpha^2}$. Given $e = \frac{1}{\sqrt{3}}$, we have $e^2 = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$. So, $\frac{1}{3} = 1 - \frac{\beta^2}{\alpha^2}$. Rearranging this, we find $\frac{\beta^2}{\alpha^2} = 1 - \frac{1}{3} = \frac{2}{3}$.

2. Eccentricity of the hyperbola: For a hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$, the eccentricity $e_H$ is given by the relation $e_H^2 = 1 + \frac{\beta^2}{\alpha^2}$.

3. Calculate $e_H^2$: Substitute the value of $\frac{\beta^2}{\alpha^2}$ found from the ellipse's eccentricity into the hyperbola's eccentricity formula: $e_H^2 = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}$.

The value of $e_H^2$ is $\frac{5}{3}$.