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

Step 1: Use the point on the ellipse to get a relation between \( a \) and \( b \). The equation of the ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
The point \( \left( \sqrt{3}, \frac{1}{2} \right) \) lies on the ellipse, so \[ \frac{(\sqrt{3})^2}{a^2} + \frac{\left(\frac{1}{2}\right)^2}{b^2} = 1 \implies \frac{3}{a^2} + \frac{1}{4b^2} = 1. \] Step 2: Use the given product of focal distances to get another relation between \( a \) and \( e \), and hence \( a \) and \( b \). The focal distances of a point \( (x, y) \) on the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) are \( a - ex \) and \( a + ex \). Their product is given as \( \frac{7}{4} \), so \[ (a - ex)(a + ex) = a^2 - e^2 x^2 = \frac{7}{4}. \] We also know that \( e^2 = 1 - \frac{b^2}{a^2} \), so \( b^2 = a^2 (1 - e^2) \). Substituting \( x = \sqrt{3} \), we get \[ a^2 - e^2 (\sqrt{3})^2 = \frac{7}{4} \implies a^2 - 3e^2 = \frac{7}{4}. \] Replacing \( e^2 = 1 - \frac{b^2}{a^2} \), we get \[ a^2 - 3 \left(1 - \frac{b^2}{a^2}\right) = \frac{7}{4} \implies a^2 - 3 + \frac{3b^2}{a^2} = \frac{7}{4} \implies a^2 + \frac{3b^2}{a^2} = \frac{19}{4}. \] Step 3: Solve the system of equations for \( a^2 \) and \( b^2 \). From the first equation, \( \frac{3}{a^2} + \frac{1}{4b^2} = 1 \), we get \( \frac{1}{4b^2} = 1 - \frac{3}{a^2} = \frac{a^2 - 3}{a^2} \), so \[ 4b^2 = \frac{a^2}{a^2 - 3} \] and \[ b^2 = \frac{a^2}{4(a^2 - 3)}. \] Substituting this into the second equation: \[ a^2 + \frac{3}{a^2} \left( \frac{a^2}{4(a^2 - 3)} \right) = \frac{19}{4} \implies a^2 + \frac{3}{4(a^2 - 3)} = \frac{19}{4}. \] Multiplying by \( 4(a^2 - 3) \) gives \[ 4a^2 (a^2 - 3) + 3 = 19(a^2 - 3) \implies 4a^4 - 12a^2 + 3 = 19a^2 - 57 \implies 4a^4 - 31a^2 + 60 = 0. \] Let \( u = a^2 \). Then \( 4u^2 - 31u + 60 = 0 \). Solving for \( u \) gives \[ u = \frac{31 \pm \sqrt{31^2 - 4(4)(60)}}{8} = \frac{31 \pm \sqrt{961 - 960}}{8} = \frac{31 \pm 1}{8}. \] So \[ a^2 = \frac{32}{8} = 4 \quad \text{or} \quad a^2 = \frac{30}{8} = \frac{15}{4}. \] If \( a^2 = 4 \), then \( b^2 = \frac{4}{4(4 - 3)} = 1 \). If \( a^2 = \frac{15}{4} \), then \[ b^2 = \frac{15/4}{4(15/4 - 3)} = \frac{15/4}{4(3/4)} = \frac{15/4}{3} = \frac{5}{4}. \] Step 4: Calculate the eccentricities and their absolute difference.
Case 1: \( a^2 = 4 \), \( b^2 = 1 \). Then \[ e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{1}{4} = \frac{3}{4}, \] so \( e = \frac{\sqrt{3}}{2} \). Case 2: \( a^2 = \frac{15}{4} \), \( b^2 = \frac{5}{4} \). Then \[ e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{5/4}{15/4} = 1 - \frac{1}{3} = \frac{2}{3}, \] so \( e = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3} \). The absolute difference of the eccentricities is \[ \left| \frac{\sqrt{3}}{2} - \frac{\sqrt{6}}{3} \right| = \left| \frac{3\sqrt{3} - 2\sqrt{6}}{6} \right| = \frac{|3\sqrt{3} - 2\sqrt{6}|}{6} = \frac{2\sqrt{6} - 3\sqrt{3}}{6} = \] \[ \frac{2\sqrt{2}\sqrt{3} - 3\sqrt{3}}{6} = \frac{\sqrt{3}(2\sqrt{2} - 3)}{6} = \frac{3-2\sqrt{2}}{-2\sqrt{3}}. \] The answer becomes \[ \frac{3-2\sqrt{2}}{2\sqrt{3}}, \] after multiplying with -1.
Final Answer: The correct answer is (3) \( \frac{3-2\sqrt{2}}{2\sqrt{3}} \).