Question

Let \( e_1 \) be the eccentricity of the hyperbola $$ \frac{x^2}{16} - \frac{y^2}{9} = 1 $$ and \( e_2 \) be the eccentricity of the ellipse $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b, $$ which passes through the foci of the hyperbola. If \( e_1 e_2 = 1 \), then the length of the chord of the ellipse parallel to the x-axis and passing through (0, 2) is: 

• A. \( 4\sqrt{5} \)
• B. \( \frac{8\sqrt{5}}{3} \)
• C. \( \frac{10\sqrt{5}}{3} \)
• D. \( 3\sqrt{5} \)

Answer 1

The Correct Option is C

To resolve the problem, we will first determine the eccentricities of the provided hyperbola and ellipse. Subsequently, we will utilize their properties to compute the length of the ellipse's chord.

1. Determine the hyperbola's eccentricity:
   • The hyperbola's equation is \( \frac{x^2}{16} - \frac{y^2}{9} = 1 \).
   • The standard hyperbola form is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), with \( a^2 = 16 \) and \( b^2 = 9 \).
   • The eccentricity \( e_1 \) of a hyperbola is calculated using \( e_1 = \sqrt{1 + \frac{b^2}{a^2}} \).
   • Upon calculation, \( e_1 = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \).
2. Identify the parameters for the ellipse passing through the hyperbola's foci:
   • The hyperbola's foci are located at \( (\pm ae_1, 0) \), where \( ae_1 = 4 \times \frac{5}{4} = 5 \).
   • Since the ellipse passes through \( (5, 0) \), substituting this point into the ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) yields:
   • \( \frac{25}{a^2} + \frac{0}{b^2} = 1 \Rightarrow a^2 = 25 \).
3. Calculate the ellipse's eccentricity:
   • The eccentricity \( e_2 \) of an ellipse is given by \( e_2 = \sqrt{1 - \frac{b^2}{a^2}} \).
   • The condition \( e_1 e_2 = 1 \) implies \( \frac{5}{4} e_2 = 1 \Rightarrow e_2 = \frac{4}{5} \).
   • Using \( e_2 = \sqrt{1 - \frac{b^2}{a^2}} \) and substituting known values: \( \left(\frac{4}{5}\right)^2 = 1 - \frac{b^2}{25} \).
   • This simplifies to \( \frac{16}{25} = 1 - \frac{b^2}{25} \), resulting in \( b^2 = 9 \).
4. Determine the length of the ellipse's chord through \( (0,2) \) parallel to the x-axis:
   • The ellipse's equation is \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \).
   • Substituting \( y = 2 \) results in: \( \frac{x^2}{25} + \frac{4}{9} = 1 \Rightarrow \frac{x^2}{25} = \frac{5}{9} \).
   • Solving for \( x^2 \): \( x^2 = \frac{25 \times 5}{9} = \frac{125}{9} \).
   • The chord length is \( 2x \), calculated as: \( 2 \times \sqrt{\frac{125}{9}} = \frac{10\sqrt{5}}{3} \).

Therefore, the length of the ellipse's chord parallel to the x-axis and passing through (0, 2) is \( \frac{10\sqrt{5}}{3} \).