Determine the point on the curve 2y=2x^2 which is nearest to the point (2,0)
Please include step by step calculations if possible... as that will help mem understand the problem better.
If I draw circles with (2,0) as center, the first one will hit the parabola tangent to the parabola, in other words with the same slope. Therefore draw a line from (2,0) tangent to the parabola. Then the radius to that point will be perpendicular to the tangent.
find slope of parabola:
dy/dx = 2x = m of parabola
slope of our radius line m' = -1/m = -1/(2x)
Our radius line goes through (2,0) so its slope is m' = -1/(2*2) = -1/4
so radius line is y = -(1/4)x +b
goes through (2,0) so
0 = -1/2 + b
b = 1/2
so
y = -(1/4) x + 1/2
Where does that radius line hit the parabola y = x^2?
y = x^2 = -x/4 + 1/2
x^2 + x/4 - 1/2 = 0
4 x^2 + x - 2 = 0
x = [ -1 +/- sqrt(1 + 32) ]/8
we want the first quadrant solution (do a sketch)
x = .593
they y = x^2 = .352
so the two points are
(2,0) and (.593 , .352 )
find the distance between those two points using
d^2 = (X2-X1)^2 + (Y2-Y1)^2