S = (-1,1) is the focus, \( 2x - 3y + 1 = 0 \) is the directrix corresponding to S and \( \frac{1}{2} \) is the eccentricity of an ellipse. If \( (a,b) \) is the centre of the ellipse, then \( 3a + 2b \) is:
S = (-1,1) is the focus, \( 2x - 3y + 1 = 0 \) is the directrix corresponding to S and \( \frac{1}{2} \) is the eccentricity of an ellipse. If \( (a,b) \) is the centre of the ellipse, then \( 3a + 2b \) is:
Show Hint
- \( \frac{30}{13} \)
- \( \frac{4}{13} \)
- \( -1 \)
\( 0 \)
The Correct Option is C
Solution and Explanation
Step 1: Using the Formula for the Centre of the Ellipse
The formula for the centre of an ellipse given a focus \( S(h,k) \), directrix \( Ax + By + C = 0 \), and eccentricity \( e \) is: \[ \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}} = e \times \text{distance of the centre from the focus}. \] Here, we have: - Focus \( S(-1,1) \). - Directrix: \( 2x - 3y + 1 = 0 \). - Eccentricity: \( e = \frac{1}{2} \). Using the formula for the centre of the ellipse: \[ (a,b) = \left(\frac{-1 + \lambda 2}{1 + \lambda^2}, \frac{1 + \lambda (-3)}{1 + \lambda^2} \right). \] Substituting \( e = \frac{1}{2} \), solving for \( a, b \), and substituting in \( 3a + 2b \): \[ 3a + 2b = -1. \]
Step 2: Conclusion
Thus, the final answer is: \[ \boxed{-1}. \]
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