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प्रश्न
The point on the curve x2 = 2y which is nearest to the point (0, 5) is ______.
पर्याय
(`2sqrt2`,4)
(`2sqrt2`,0)
(0, 0)
(2, 2)
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उत्तर
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`
⇒ 2y - 8 = 0 ∴ y = 4
From equation (2), x2 = 2 x 4 = 8 ∴ x = 2`sqrt2`
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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