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?

1 Answer
May 24, 2016

Answer:

If m is the slope of line joining the origin O and P (x, #sqrt x)#, the the coordinates of P become #(m^2, m), m >=0#. This is a parametric form of the coordinates, for the curve #y = sqrt x#..

Explanation:

Here, # x >= 0#. The point P is #( x, sqrt x )#.

If m is slope of the line joining the origin #O (0, 0) and P (x, sqrt x ),#
#m = sqrt x / x =sqrt x#.

Now, if m is used as a parameter, P is #(m^2, m), m>=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#.