Question

Let the product of the focal distances of the point $$ \left( \sqrt{3}, \frac{1}{2} \right) $$ on the ellipse $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad (a > b), $$ be $ \frac{7}{4} $. Then the absolute difference of the eccentricities of two such ellipses is:

• A. \( \frac{3 - 2\sqrt{2}}{3\sqrt{2}} \)
• B. \( \frac{1 - \sqrt{3}}{\sqrt{2}} \)
• C. \( \frac{3 - 2\sqrt{2}}{2\sqrt{3}} \)
• D. \( \frac{1 - 2\sqrt{2}}{\sqrt{3}} \)

Hint:

When solving for the eccentricities of ellipses, remember that the focal distance formula involves the terms \( a^2 - b^2 \), and the eccentricity \( e = \frac{\sqrt{a^2 - b^2}}{a} \). This formula helps find the difference between eccentricities when dealing with different ellipses.

Answer 1

The Correct Option is C

Given the following equations:

\( (a + e\sqrt{3})(a - e\sqrt{3}) = \frac{7}{4} ……{i} \)

\( \frac{3}{a^2} + \frac{1}{4b^2} = 1 …….{ii} \)

\( b^2 = a^2(1 - e^2) ……….{iii} \)

Step 1: Solve the equation from (i)

From equation (i):

\( (a + e\sqrt{3})(a - e\sqrt{3}) = a^2 - e^2 \cdot 3 = \frac{7}{4} \) \[ a^2 - 3e^2 = \frac{7}{4}. \]

Step 2: Use equations (ii) and (iii)

From equation (ii), we have:

\[ \frac{3}{a^2} + \frac{1}{4b^2} = 1 \] Simplifying gives a relationship between \( a \) and \( b \). Using equation (iii), we substitute \( b^2 = a^2(1 - e^2) \) to solve for \( e \).

Step 3: Solve the quadratic equation for \( e \)

Substituting into the equation gives us: \[ 12e^4 - 17e^2 + 6 = 0. \] Factoring this quadratic equation: \[ (3e^2 - 2)(4e^2 - 3) = 0 \] This yields: \[ e = \sqrt{\frac{2}{3}} \quad \text{or} \quad e = \sqrt{\frac{3}{4}}. \]

Step 4: Calculate the difference

The difference is: \[ \frac{\sqrt{3}}{2} - \sqrt{\frac{2}{3}} = \frac{3 - 2\sqrt{2}}{2\sqrt{3}}. \]

Answer:

The correct option is (3). The final difference is \( \frac{3 - 2\sqrt{2}}{2\sqrt{3}} \).