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Question

The co-ordinate of a point on the parabola y=x2+7x+2 which is nearest to the straight line y=3x−3 is

A
(2,8)
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B
(2,8)
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C
(2,8)
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D
(2,8)
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Solution

The correct option is C (2,8)
Let P be the point on the line closest to the parabola and let the shortest distance line the ray P meet the parabola in Q.
Then PQ, is the shortest distance between the line and the parabola.
RQ must be perpendicular to the given line and must be normal to the parabola at Q.
Let the equation of the line PQ be =13x+k (Since PQ is to the given line)
Let the co-ordinates of Q be (x1,y1)
Slope of the tangent to the parabola at x=x1 is given by dydx=2x1+7
Slope of the normal at x=x1 is 1(2x1+7)
12x1+7=13x1=2
y1=x12+7x1+2=414+2=8
Q is (2,8)

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