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suppose that,we have given following parabolic arc $\sqrt{x}+\sqrt{y}=\sqrt{a}$ we are trying to find shortest distance from origin to this line, i think that if we rewrite it as $\sqrt{y}=\sqrt{a}-\sqrt{x}$, then

$y=a-2\sqrt{ax}-x$ or if we rearrange terms,we get $y+a-\sqrt{ax}-x=0$ distance from origin $(0,0)$ to line $Ax+by+c=0$ is equal $D=(A\cdot0+b\cdot0+c)/(\sqrt{A^2+b^2})$ so in my case what would be minimum distance? please help me

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Notice that $A$-$B$ looks different from $A-B$. The latter is achieved by enclosing the whole thing within dollar signs so that the minus sign is set according to the standard typesetting conventions for minus signs. Similarly if you want $\sqrt{ABCD}$, enclose "ABCD" within curly braces rather than writing $\sqrt(ABCD)$. –  Michael Hardy Feb 19 '12 at 18:16

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One asks that the vector $(x,y)$ is orthogonal to the tangent at $(x,y)$ of the curve. Thus, assume that $(x,y)$ and $(x+h,y+k)$ are on the curve and that $h,k\to0$. Then $$ \sqrt{x}+\sqrt{y}=\sqrt{a}=\sqrt{x+h}+\sqrt{y+k}=\sqrt{x}\sqrt{1+h/x}+\sqrt{y}\sqrt{1+k/y} $$ hence $$ \sqrt{x+h}+\sqrt{y+k}=\sqrt{x}+h/(2\sqrt{x})+o(h)+\sqrt{y}+k/(2\sqrt{y})+o(k). $$ In particular, $h/\sqrt{x}\sim-k/\sqrt{y}$ hence the tangent vector is $(\sqrt{x},-\sqrt{y})$.

The tangent is orthogonal to $(x,y)$ if and only if $x\sqrt{x}-y\sqrt{y}=0$, that is, $x=y$, that is $x=y=a/4$. The shortest distance is $\sqrt{x^2+y^2}=\sqrt2a/4$.

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