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प्रश्न
Find the point on the curve y2 = 4x, which is nearest to the point (2, 1).
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उत्तर १
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).
उत्तर २
Suppose the required point on the curve is K(p, q) and the given point is A(2, 1).
∴ q2 = 4p ...(i)
Then, AK = `sqrt((p - 2)^2 + (q - 1)^2`
⇒ S = `sqrt((q^2/4 - 2)^2 + (q - 1)^2` ...[Let S = AK from (i) `p = q^2/4`]
Then, S2 = `(q^2/4 - 2)^2 + (q - 1)^2`
To find nearest point let us suppose S2 = T
Then, T = `(q^2/4 - 2)^2 + (q - 1)^2`
⇒ T' = `2(q^2/4 - 2) xx (2q)/4 + 2(q - 1)`
For critical points T' = 0
⇒ `2(q^2/4 - 2) xx (2q)/4 + 2(q - 1) = 0`
⇒ `(q^2/4 - 2q) + (2q - 2) = 0`
⇒ q3 = 8
⇒ q = 2 ...(ii)
To find the maxima or minima
T" = `(3q^2)/4 - 2 xx 2`
= `(3q^2)/4`
⇒ T" at 2 = `(3 xx 2 xx 2)/4`
= 3 > 0
Therefore, T is least
From (i) q2 = 4p
⇒ 22 = 4p
⇒ p = 1
Therefore, the required point is K(1, 2).
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