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Find the distance between the origin and the normal to the curve $y =e^{2x}+x^2$ drawn at the point $x =0$

After solving , I got the equation of normal is : $x+2y=2$

Please guide further, thanks.

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Hint:

Now use the formula of distance from a given point:

$|\dfrac{Ax+By+C}{ \sqrt{A^2+B^2}}|$ is the distance from a point $(x,y$) for a line $Ax_1+By_1+C=0$

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  • $\begingroup$ thanks i got the answer $\frac{2}{\sqrt{5}}$ thanks once again to all of you guys.. $\endgroup$ – sultan May 9 '13 at 8:56
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There's a formula for the distance of a point from a line $y=mx+c$, you may use it:

$$\text{distance}=\left|\frac{y+mx-c}{\sqrt{1+m^2}}\right|$$

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Hint: The distance from the origin to $(x,y)$ is $\sqrt{x^2+y^2}$. But you know that $x=2-2y$ for points on your normal.

So we want to minimize $\sqrt{(2-2y)^2+y^2}$, or equivalently to minimize $$(2-2y)^2+y^2.$$ Now it's your turn.

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