Question

find whether the point (2,10/3) lie inside or outside the parabola 2x-x^2+3y=0.

Answer 1

Let 2x-x²+3y=0
or
f(x)=y=(1/3)x²-(2/3)x
It is a curve concave upwards.

To find out if P0(x0,y0)is above or below, calculate
h=f(x0)-y0
if h>0, P0 is below the parabola (outside), and if h<0, P0 is above the parabola (inside in this case).

For the given problem,
f(x0)-y0
=(1/3)2²-(2/3)2 -3/10
=4/3-4/3-3/10
=-3/10
<0, therefore the point is above the parabola (concave upwards), so it lies inside.