The point on the curve `x^2=2y`which is nearest to the point (0, 5) is(A) `(2sqrt(2),4)` (B) `(2sqrt(2),0)` (C) (0, 0) (D) (2, 2)

Options

A. `(2sqrt2, 4)`
B. `(2 sqrt2, 0)`
C. ``(0, 0)`
D. `(2, 2)`

Answer

A

Explanation

Let d be the distance of point (x, y) on ` x^(2) = 2y` from the point(0, 5),
`d = sqrt((x-0)^(2)+(y-5)^(2)) = sqrt(x^(2)+(y-5)^(2))`
` = sqrt(2y+(y-5)^(2))` ....(1)
` rArr d = sqrt(2y+y^(2)-10y+25)`
` = sqrt(y^(2)-8y+4^(2)+9)=sqrt((y-4)^(2)+9)`
d will be minimum when `(y-4)^(2)=0 or y= 4`
`:. y = 4 rArr x^(2)=2 xx 4`
` rArr x = pm sqrt8 = pm 2 sqrt2`
`:." Points "(2sqrt2,4) and (-2sqrt2,4)` are the points on the curve, at minimum distance from the point (0, 5).