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

The objective is to compute the absolute difference between the eccentricities of two potential ellipses, given specific information about one ellipse.

The provided ellipse is defined by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a > b\). The relationship between focal distances and ellipse properties is utilized. It is stated that for the point \((\sqrt{3}, \frac{1}{2})\), the product of its focal distances is \(\frac{7}{4}\).

The general formula for the product of focal distances of a point \((x_1, y_1)\) on an ellipse is:

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

Substituting \(x_1 = \sqrt{3}\) into this formula yields:

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

This simplifies to:

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

Using the relations \(c^2 = a^2 - b^2\) and \(e = \frac{c}{a}\), the eccentricity is expressed as:

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

The eccentricity \('e'\) is solved for two values, derived from the positive square roots:

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

Two forms for \(b^2\) are considered:

• First ellipse: \(b_1 = \frac{\sqrt{7a^2}}{\sqrt{4(a^2 - 3)}}\)
• Second ellipse: The second form of \(b^2\), represented by a negative square root, does not yield a distinct ellipse under the modulus consideration for eccentricity.

The absolute difference between \(e_1\) and \(e_2\) is then calculated:

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

This result matches the option \(\frac{3 - 2\sqrt{2}}{2\sqrt{3}}\), confirming it as the correct answer.