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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If \(a\) and \(c\) are positive real numbers and the ellipse \(\frac{{{x^2}}}{{4{c^2}}} + \frac{{{y^2}}}{{{c^2}}} = 1\) has four distinct points in common with the circle \(x^2 + y^2 = 9a^2\) , then
- A \(9ac -9a^2 - 2c^2 <0\)
- B \(6ac + 9a^2 - 2c^2 < 0\)
- C \(9ac -9a^2 -2c^2 > 0\)
- D \(6ac +9a^2 - 2c^2 >0\)
Answer & Solution
Correct Answer
(C) \(9ac -9a^2 -2c^2 > 0\)
Step-by-step Solution
Detailed explanation
Radius \(=3a\) Length of major axis \(=4c\) Now, (radius)<(Half of the length of major axis) \(3a < 2c\) \(9{a^2} < 4{c^2}\) \(9ac - 9{a^2} > 9ac - 4{c^2}\) \(9ac - 9{a^2} - 2{c^2} > 9ac - 6{c^2}\,\,\,\,\,\,\,\,......\left( i \right)\) Again \(3a < 2c\)…
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