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A line is drawn through a fixed point (a,b) to meet the $X$-axis and $Y$-axis at $P$ and $Q$ respectively. Show that the minimum values of $PQ$, $OP+OQ$, and $OP\cdot OQ$ are respectively $(a^{2/3}+b^{2/3})^{3/2}$, $(\sqrt{a}+\sqrt{b})^2$, and $4ab$.

I set up the solution like this: $\frac{Q-b}{0-a}=\frac{b-0}{a-P}$. Is it correct?

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    what is P in that ratio??2012-06-24
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    I assume the line intersects x and y axes at P and Q respectively, so the ratio is supposed to be the slope of the line. This is how I interpreted the question. I am not sure that I am right.2012-06-24
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    use slope intercept form ... it is given below in my answer.2012-06-24
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    For some reason that i don't know [Wolframalpha](http://www.wolframalpha.com/input/?i=minimize+sqrt%28x^2+%2B+%28ax%2F%28x-b%29+%29^2%29+for+x) seems to say that there is no global minimum ... are you sure that question is complete.2012-06-24
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    Have you noticed the power $2/3$?....It means $a,b\ge0$.2012-06-24
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    experimentX, the question is complete. I typed it verbatim. It's from the Calculus book written by George Thomas of MIT. I also found this exact question in "A course in pure mathematics" by G.H. Hardy. I solved the value of PQ; it's exactly as required, but I can never get the result for OP.2012-06-25

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