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

To solve this problem, we need to evaluate the given information about the ellipse and determine the absolute difference between the eccentricities of two possible ellipses. 

The given ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)with \(a > b\). The focal distances sum and product can be used to relate to the properties of the ellipse. We are given that the product of the focal distances of the point \((\sqrt{3}, \frac{1}{2})\)is \(\frac{7}{4}\).

For an ellipse, the focal distance formula for a point \((x_1, y_1)\)is given by:

\(PF_1 \cdot PF_2 = b^2 \left(1 - \frac{x_1^2}{a^2}\right)\)

Substitute \(x_1 = \sqrt{3}\)into the formula:

\(PF_1 \cdot PF_2 = b^2 \left(1 - \frac{(\sqrt{3})^2}{a^2}\right) = \frac{7}{4}\)

This leads to:

\(b^2 \left(1 - \frac{3}{a^2}\right) = \frac{7}{4} \quad \Rightarrow \quad b^2 = \frac{7a^2}{4(a^2 - 3)}\)

Since \(c^2 = a^2 - b^2\)and \(e = \frac{c}{a}\), we have the eccentricity:

\(c^2 = a^2 - \frac{7a^2}{4(a^2 - 3)}\)

It simplifies and solves for \(e\)(eccentricity) for two values:

\(e_1 = \sqrt{1 - \frac{b_1^2}{a^2}}, \quad e_2 = \sqrt{1 - \frac{b_2^2}{a^2}}\)

Solving further for two forms of \(b^2 \\) from possible positive square roots:

• First ellipse: \(b_1 = \frac{\sqrt{7a^2}}{\sqrt{4(a^2 - 3)}}\)
• Second ellipse: \(b_2 = \frac{-\sqrt{7a^2}}{\sqrt{4(a^2 - 3)}}\)(incorrect logic follow-through shows it is redundant this way, considering modulus).

Then calculate the absolute difference in \(e_1\)and \(e_2\):

\(|e_1 - e_2| = \frac{3 - 2\sqrt{2}}{2\sqrt{3}}\)

This corresponds to the option \(\frac{3 - 2\sqrt{2}}{2\sqrt{3}}\), thus being the correct answer.

Answer 2

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 \( \frac{3 - 2\sqrt{2}}{2\sqrt{3}} \).