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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate axes and passing through the points \((4,-1)\) and \((-2, 2)\) is
- A \(\frac{1}{2}\)
- B \(\frac{2}{{\sqrt 5 }}\)
- C \(\frac{{\sqrt 3 }}{2}\)
- D \(\frac{{\sqrt 3 }}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{{\sqrt 3 }}{2}\)
Step-by-step Solution
Detailed explanation
Centre at \((0,0)\) \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) at point \((4,-1)\) \(\frac{{16}}{{{a^2}}} + \frac{1}{{{b^2}}} = 1\) \(16{b^2} + {a^2} = {a^2}{b^2}\,\,\,\,\,\,\,\,.......\left( i \right)\) at point \((-2,2)\)…
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