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

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KollegeApply · Accepted Answer

The point on the curve x2 = 2y which is nearest to the point (0, 5) is underline(2sqrt2,4). Explanation: Let P(x, y) be any point on the curve x2 = 2y. The given point is A (0, 5). PA2 = (x - 0)2 + (y - 5)2 = z (Let) Z = x2 + (y - 5)2 …(1) And the curve x2 = 2y …(2) Putting the value of x2 in equation (1), Z = 2y + (y - 5)2 = 2y + y2 + 25 - 10y = y2 + 25 - 8y Differentiating both sides with respect to y, (dZ)/dy =2y- 8 For highest and lowest values, (dZ)/dy = 0 $\Rightarrow$ 2y - 8 = 0 $\therefore$ y = 4 From equation (2), x2 = 2 x 4 = 8 $\therefore$ x = 2sqrt2 Differentiating both sides again with respect to y, (d^2Z)/(dy^2) = 2 = +ve Z is minimum at x = 2 sqrt2 y = 4.

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