Question
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.
Please include step by step calculations if possible... as that will help mem understand the problem better.
Answers
Damon
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
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