Question:
If the eccentricity of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1,(a > b)\) is \(\frac{2}{3}\) and its focal chord is \(3x+2y-6 = 0\), then the value of \(a^2+b^2\) is...
If the eccentricity of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1,(a > b)\) is \(\frac{2}{3}\) and its focal chord is \(3x+2y-6 = 0\), then the value of \(a^2+b^2\) is...
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A focal chord passes through a focus, which lies at (ae, 0).
Updated On: Oct 1, 2026
- \(11\)
- \(12\)
- \(13\)
- \(14\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
A focal chord is a chord that passes through a focus. For the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with \(a > b\), the foci are \((\pm ae, 0)\).
Step 2: Key Formula or Approach:
The line \(3x + 2y - 6 = 0\) meets the x-axis at \((2, 0)\), so this point is a focus: \(ae = 2\).
Step 3: Detailed Explanation:
Given \(e = \frac23\): \(a \times \frac23 = 2\), so \(a = 3\) and \(a^2 = 9\).
\(b^2 = a^2(1 - e^2) = 9\left(1 - \frac49\right) = 5\).
\[ a^2 + b^2 = 9 + 5 = 14 \]
Final Answer:
\(a^2 + b^2 = 14\), option (D). \[ \boxed{14} \]
A focal chord is a chord that passes through a focus. For the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with \(a > b\), the foci are \((\pm ae, 0)\).
Step 2: Key Formula or Approach:
The line \(3x + 2y - 6 = 0\) meets the x-axis at \((2, 0)\), so this point is a focus: \(ae = 2\).
Step 3: Detailed Explanation:
Given \(e = \frac23\): \(a \times \frac23 = 2\), so \(a = 3\) and \(a^2 = 9\).
\(b^2 = a^2(1 - e^2) = 9\left(1 - \frac49\right) = 5\).
\[ a^2 + b^2 = 9 + 5 = 14 \]
Final Answer:
\(a^2 + b^2 = 14\), option (D). \[ \boxed{14} \]
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