# A point P in the first quadrant lies on the graph of the function f(x) = sqrt x, how do you express the coordinates of P as function of the slope of the line joining P to the origin?

May 24, 2016

If m is the slope of line joining the origin O and P (x, sqrt x), the the coordinates of P become $\left({m}^{2} , m\right) , m \ge 0$. This is a parametric form of the coordinates, for the curve $y = \sqrt{x}$..

#### Explanation:

Here, $x \ge 0$. The point P is $\left(x , \sqrt{x}\right)$.

If m is slope of the line joining the origin $O \left(0 , 0\right) \mathmr{and} P \left(x , \sqrt{x}\right) ,$
$m = \frac{\sqrt{x}}{x} = \sqrt{x}$.

Now, if m is used as a parameter, P is $\left({m}^{2} , m\right) , m \ge 0$.

Thanks to Abhishek Dogra, I have now revised my answer to

remove the error pointed out by him.

The graph is a semi-parabola.

The other half is given by $y = - \sqrt{x}$.