+0

0
42
1

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
Sort:

#1
0

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

### 7 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details