The point on x^ 2 =2 y . which is closest to the point (0, 5), will be
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
The point on \(x^{2}=2 y .\) which is closest to the point (0, 5), will be
- \((2 \sqrt{2}, 0)\)
- (0, 0)
- (2, 2)
- (1, 8)
Answer
(C) (2, 2)
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