Question:
If \(a\) and \(c\) are positive real numbers and the ellipse \(\frac{x^2}{4c^2}+\frac{y^2}{c^2}=1\) has four distinct points in common with the circle \(x^2+y^2=9a^2\), then
If \(a\) and \(c\) are positive real numbers and the ellipse \(\frac{x^2}{4c^2}+\frac{y^2}{c^2}=1\) has four distinct points in common with the circle \(x^2+y^2=9a^2\), then
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The circle meets the ellipse at four points exactly when its radius lies strictly between the semi-minor and semi-major axes.
Updated On: Oct 31, 2026
- \(9ac-9a^2-2c^2\lt0\)
- \(6ac+9a^2-2c^2\lt0\)
- \(9ac-9a^2-2c^2\gt0\)
- \(6ac+9a^2-2c^2\gt0\)
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The Correct Option is C
Solution and Explanation
The ellipse has semi-axes \(2c\) (along x) and \(c\) (along y). A concentric circle of radius \(3a\) meets it at four distinct points iff \(c\lt 3a\lt 2c\).
So \((3a-c)\gt0\) and \((2c-3a)\gt0\), hence \((3a-c)(2c-3a)\gt0\).
Expanding: \(6ac-9a^2-2c^2+3ac=9ac-9a^2-2c^2\gt0\).
Answer: (3) \(9ac-9a^2-2c^2\gt0\)
So \((3a-c)\gt0\) and \((2c-3a)\gt0\), hence \((3a-c)(2c-3a)\gt0\).
Expanding: \(6ac-9a^2-2c^2+3ac=9ac-9a^2-2c^2\gt0\).
Answer: (3) \(9ac-9a^2-2c^2\gt0\)
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