The points on the curve x^ 2 =2 y , which are closest to the point (0, 5) are
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
The points on the curve \(x^{2}=2 y\), which are closest to the point (0, 5) are
- \((2,2),(-2,2)\)
- \((2 \sqrt{2}, 4),(-2 \sqrt{2}, 4)\)
- \((\sqrt{6}, 3),(-\sqrt{6}, 3)\)
- \((2 \sqrt{3}, 6),(-2 \sqrt{3}, 6)\)
Answer
(B) (2 2 , 4),(-2 2 , 4)
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