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**1. Homework Statement**

"find a point P on the parabola [tex]x^2-4y = 0[/tex] for which is minimum the distance PO/PF, where O is the origin and F the focus of the parabola"

the solution is P(0;0), coincides with the origin, this is also clear graphically.

**3. The Attempt at a Solution**

Please follow these lines:

1) i set up the function PO/PF, and i call it "y". [O(0;0), F(0;1), P(x;x²/4)]

[tex]y = \frac{\sqrt{x^2+x^4/16}}{\sqrt{x^2+(\frac{x^2}{4}-1)^2}}[/tex]

simplified:

[tex]y = \sqrt{\frac{16x^2+x^4}{x^4+8x^2+16}}[/tex]

2) now i set up the derivative:

[tex]y' = \frac{\frac{(32x+4x^3)(8x^2+x^4+16) - (16x+4x^3)(16x^2+x^4)}{(x^4+8x^2+16)^2}}{2\sqrt{\frac{16x^2+x^4}{x^4+8x^2+16}}}[/tex]

3) i put the derivative equal to zero:(some stuff simplifies)

[tex]y' = 0 = (32x+4x^3)(8x^2+x^4+16) - (16x+4x^3)(16x^2+x^4)[/tex]

solving: [tex]x^4-4x^2-32 = 0[/tex]

4) this thing should give me as result x=0, in order to have P(0;0), but it doesnt...

i checked and checked again, could you find the error inside?

thanks.

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