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The coordinates of the point on the parabola $\mathrm { y=x^2+6 x+5 }$ , which is nearest to the straight line $\mathrm { 2 x-y-5=0 }$ , is

Option: 1
$(-3,-2)$

Option: 2
$(3,2)$

Option: 3
$(2,3)$

Option: 4
$(-2,-3)$

Answers (1)

best_answer

Any point on the parabola is $\mathrm {A\left(t, t^2+6 t+5\right) or \: \: A\left(x, x^2+6 x+5\right)}$ . In order to draw the graph, we have
$\mathrm { y=x^2+6 x+5 }$

$\mathrm { or\: \: y=(x+3)^2-4 }$

$\mathrm { \Rightarrow(x+3)^2=y+4 \quad \Rightarrow \quad X^2=Y }$

$\mathrm { \therefore }$ Vertex is $\mathrm { (-3,-4) }$ , parabola cut x-axis at $\mathrm { (-1,0),(-5,0) }$ & y-axis at $\mathrm { (0,5). }$

As $\mathrm { A\left(x, x^2+6 x+5\right) }$ be any point on the parabola,

therefore let $\mathrm { S= }$ Distance from A to the line $\mathrm { 2 x-y-5=0 }$

Then, $\mathrm { S=\frac{\left|2 x-x^2-6 x-5-5\right|}{\sqrt{5}}$
$\mathrm { S=\frac{\left|x^2+4 x+10\right|}{\sqrt{5}}=\frac{(x+2)^2+6}{\sqrt{5}} }$

which will be minimum if $\mathrm { (x+2)=0 \: \: i.e., x=-2 }$

$\mathrm { \therefore(y)_{x=-2}=\left(x^2+6 x+5\right)_{x=-2}=-3 }$

$\mathrm { \therefore }$ Required point is (-2,-3)

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