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The smallest distance between the origin and a point on the parabola $y=x^2-5$ can be expressed as $\sqrt{a}/b$, where $a$ is not divisible by the square of any prime. Find $a+b$.

Guest Mar 6, 2018
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The distance from a point (x, y) to the origin is given by \(\displaystyle D = \sqrt{x^{2}+y^{2}}.\)

So, if

 \(\displaystyle y=x^{2}-5,\text{ then}\\D=\sqrt{x^{2}+(x^{2}-5)^{2}}=\sqrt{x^{4}-9x^{2}+25}.\)

 

Now complete the square, (add and subtract (9/2)^2),

 

\(\displaystyle D = \sqrt{x^{4}-9x^{2}+(81/4)+(100/4)-(81/4)}\\=\sqrt{\{x^{2}-(9/2)\}^{2}+(19/4)}.\)

 

This will take its smallest value 

\(\displaystyle \sqrt{19/4}=\sqrt{19}/2, \text{ when }x^{2}=9/2.\)

 

Tiggsy

Guest Mar 7, 2018

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