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:
- \( 4\sqrt{5} \)
- \( \frac{8\sqrt{5}}{3} \)
- \( \frac{10\sqrt{5}}{3} \)
- \( 3\sqrt{5} \)
The Correct Option is C
Approach Solution - 1
To solve the problem, we first need to find the eccentricities of the given hyperbola and ellipse and then use their properties to calculate the length of the chord of the ellipse.
- Find the eccentricity of the hyperbola:
- The equation of the hyperbola is \(\frac{x^2}{16} - \frac{y^2}{9} = 1\).
- The standard form for a hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a^2 = 16\) and \(b^2 = 9\).
- The eccentricity \(e_1\) of a hyperbola is given by \(e_1 = \sqrt{1 + \frac{b^2}{a^2}}\).
- Calculating \(e_1\), we get: \(e_1 = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}\).
- Find the parameters for the ellipse that passes through the foci of the hyperbola:
- The foci of the hyperbola are located at \((\pm ae_1, 0)\), where \(ae_1 = 4 \times \frac{5}{4} = 5\).
- The ellipse passes through \((5, 0)\), so substituting this point into the ellipse's equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we get:
- \(\frac{25}{a^2} + \frac{0}{b^2} = 1 \Rightarrow a^2 = 25\)
- Find the eccentricity of the ellipse:
- The eccentricity \(e_2\) of an ellipse is given by \(e_2 = \sqrt{1 - \frac{b^2}{a^2}}\).
- The condition \(e_1 e_2 = 1\) gives us: \(\frac{5}{4} e_2 = 1 \Rightarrow e_2 = \frac{4}{5}\).
- Using \(e_2 = \sqrt{1 - \frac{b^2}{a^2}}\), we substitute the known values: \(\left(\frac{4}{5}\right)^2 = 1 - \frac{b^2}{25}\).
- This results in: \(\frac{16}{25} = 1 - \frac{b^2}{25}\), which simplifies to \(b^2 = 9\).
- Find the length of the chord of the ellipse through \((0,2)\) parallel to the x-axis:
- The equation of the ellipse is \(\frac{x^2}{25} + \frac{y^2}{9} = 1\).
- Substituting \(y = 2\), we have: \(\frac{x^2}{25} + \frac{4}{9} = 1 \Rightarrow \frac{x^2}{25} = \frac{5}{9}\).
- Solving for \(x^2\), we get \(x^2 = \frac{25 \times 5}{9} = \frac{125}{9}\).
- The length of the chord is given by \(2x\), which is: \(2 \times \sqrt{\frac{125}{9}} = \frac{10\sqrt{5}}{3}\).
Hence, the length of the chord of the ellipse parallel to the x-axis and passing through (0, 2) is \(\frac{10\sqrt{5}}{3}\).
Approach Solution -2
Given:
\(\frac{x^2}{16} + \frac{y^2}{9} = 1 \implies e_1 = \sqrt{1 - \frac{9}{16}} = \frac{5}{4}.\)
For the ellipse:
\(e_1 e_2 = 1 \implies e_2 = \frac{4}{5}.\)
The ellipse passes through \((\pm 5, 0)\), so \(a = 5\) and \(b = 3\):
\(\frac{x^2}{25} + \frac{y^2}{9} = 1.\)
The length of the chord parallel to the \(x\)-axis and passing through \((0, 2)\) is given by:
\(L = 2a \sqrt{1 - \frac{y^2}{b^2}} = 2 \times 5 \times \sqrt{1 - \frac{4}{9}} = 10 \sqrt{\frac{5}{9}} = \frac{10 \sqrt{5}}{3}.\)
The Correct answer is: \( \frac{10\sqrt{5}}{3} \)
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