Question

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

Options

• `(2sqrt(2), 4)`

• `(2sqrt(2), 0)`

• `(0, 0)`

• `(2, 2)`

Answer 1

`(2sqrt(2), 4)`

Explanation:

Let `P(x, y)` be a point on the curve

The after point is A(0, 5)

Z = PA² = `(x - 0)^2 + (y - 5)^2`

Z = `x^2 + (y - 5)^2 = 2y + (y - 5)^2`  ......`[because x^2 = 2y]`

= `y^2 - 8y + 25`

∴ `(dz)/(dy) = 2y - 8, (d^2z)/(dy^2)` = 2 = positive

`(dz)/(dy) ⇒ y` = 4

`(d^2z)/(dy^2)` = positive, Z is minimum

∴ `x^2 = 2y = 2 xx 4 = 8`

∴ `x = 2sqrt(2)`

⇒ Z is minimum at `(2sqrt(2), 4)`