Question 64 of 82intermediate🔧 ApplyMCQ1 mark

The point on the curve x² = 2y which is nearest to the point (0, 5) is

Exercise: EXERCISE 6.3 | Q: 27 | (Page 31)
For More Understanding

Explanation

To find the nearest point on the curve x² = 2y to the point (0, 5), we minimize the distance squared. Let (x, y) be a point on the curve, so y = x²/2. The distance squared from (x, y) to (0, 5) is D² = x² + (y - 5)². Substituting y = x²/2 and simplifying, we get D² = y² - 8y + 25. Taking derivative and setting to zero: d(D²)/dy = 2y - 8 = 0, giving y = 4. The second derivative is positive (2 > 0), confirming minimum. When y = 4, x² = 2(4) = 8, so x = ±22\sqrt{2}. The point (22\sqrt{2}, 4) is on the curve and gives minimum distance of 3 units to (0, 5).

Solution Steps

  1. Step 1: Let (x, y) be a point on curve x² = 2y, so y = x²/2

  2. Step 2: Distance squared from (x, y) to (0, 5): D² = x² + (y - 5)²

  3. Step 3: Substitute y = x²/2: D² = x² + (x²/2 - 5)² = y² - 8y + 25

  4. Step 4: Minimize: d(D²)/dy = 2y - 8 = 0, gives y = 4

  5. Step 5: Verify minimum: d²(D²)/dy² = 2 > 0

  6. Step 6: Find x: x² = 2(4) = 8, so x = 22\sqrt{2}

  7. Step 7: Point is (22\sqrt{2}, 4), which matches option (A)

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