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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the points of the conic $y=x^2+7x+2$ which is nearest to the line $3x-y-3=0$

$\begin{array}{1 1}(A)\;(2,8) \\(B)\;(2,7) \\(C)\;(-2,-8) \\(D)\;(4,5) \end{array}$

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1 Answer

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Any point P on the conic is $(x,x^2+7x+2)$
Its distance from $3x-y-3=0$ is
$D= \large\frac{|3x-(x^2+7x+2)-3|}{\sqrt {9+1}}$
$\qquad= \large\frac{x^2+4x+5}{\sqrt {10}}$
$\large\frac{dD}{dx}=\frac{2x+4}{\sqrt {10}}$
$\large\frac{d^2D}{dx^2}=\frac{2}{\sqrt {10}} > 0$
$\therefore$ D is least when $2x+4=0$
or $x=-2$
Required point is $(-2,-8)$
Hence D is the correct answer.
answered Apr 9, 2014 by meena.p
 

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