CUET · MATHS · PYQ PAPER 2023
The point on the curve \(x^2=2 y\) in the I quadrant which is nearest to the point (0, 5) is:
- A \((2 \sqrt{2}, 0)\)
- B (0, 0)
- C \((2 \sqrt{2}, 4)\)
- D (2, 2)
Answer & Solution
Correct Answer
(C) \((2 \sqrt{2}, 4)\)
Step-by-step Solution
Detailed explanation
Let the point on the curve be \((x, y)\). Given \(x^2=2y \Rightarrow y=x^2/2\). Distance squared from \((x, y)\) to \((0, 5)\) is \(D^2 = (x-0)^2 + (y-5)^2\). \(D^2 = x^2 + (x^2/2 - 5)^2\). To minimize \(D^2\), we find the derivative with respect to \(x\) and set it to zero:…
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