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Question

27. The point on the curve x2- 2y which is nearest to the point (0, 5) is(A) (2V2,4) (B) (2^2,0) (C) (o.0) (D) (2,2)

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Solution

The given equation is given as,

x 2 =2y(1)

It is given that the curve is nearest to the point ( 0,5 ).

Consider point P( x,y ) as any point on the curve.

The formula for the distance between two points is,

d= ( x x 1 ) 2 + ( y y 1 ) 2

Substitute 0 for x 1 and 5 for y 1 in the above equation.

d= ( x0 ) 2 + ( y5 ) 2 d 2 = x 2 + ( y5 ) 2

Substitute the value x 2 =2y from equation (1)

d 2 =2y+ ( y5 ) 2 d 2 =2y+ y 2 +2510y d 2 = y 2 8y+25

Consider,

d 2 =M M= y 2 8y+25

For maxima or minima,

dM dy =0 dM dy =2y8 2y8=0 y=4

Also

d 2 M d y 2 =2 =+ve

Hence, M is minimum and d is minimum at y=4.

From equation (1)

x 2 =8 x=±2 2

Therefore, ( 2 2 ,4 ) and ( 2 2 ,4 ) are two points on curve nearest to (0,5).

Thus, the correct option is (A).


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