Question

Find the point on the curve y² = 4x, which is nearest to the point (2, 1).

Answer 1

Let P(x, y) be a point on the curve y² = 4x

which is nearest to the point A(2, 1).

Let D = AP

∴ D² = (x – 2)² + (y – 1)²

Put x = `y^2/4`

∴ D² = `(y^2/4 - 2)^2 + (y - 1)^2`

Differentiating w.r.t. y,

`2D*(dD)/dy = 2(y^2/4 - 2)((2y)/4) + 2(y - 1)`

∴ `2D (dD)/dy = y(y^2/4 - 2) + 2y - 2`

`2D*(dD)/dy = y^3/4 - 2y + 2y - 2`

∴ `D*(dD)/dy = y^3/8 - 1`

`\implies (dD)/dy = 1/D(y^2/8 - 1)`

For extreme value of D, put

`(dD)/dy` = 0   ...(1)

∴ `1/D(y^3/8 - 1)` = 0

But D > 0

`\implies y^3/8 - 1` = 0

∴ y³ = 8

∴ y = 2

Now `D*(dD)/dy = y^3/8 - 1`

Again differentiating w.r.t. y,

`D(d^2D)/(dy^2) + ((dD)/dy)^2 = 3/8y^2`

But `(dD)/dy` = 0   ...[From (1)]

∴ `D*(d^2D)/(dy^2) = 3/8y^2`

∴ `(d^2D)/(dy^2) = 1/D*3/8y^2`

∴ `((d^2D)/dy^2)_(y = 2) = 1/D*3/8 xx 4`

= `3/(2D) > 0`   ...(∵ D > 0)

∴ D is minimum when y = 2

∴ y² = 4x gives x = 1

∴ The point (1, 2) is the point nearest to (2, 1).