Find the minimum distance from the parabola

Question

Find the minimum distance from the parabola x-y^2=0 to the point {(0,3)}. 

 

Minimum distance=______?

Answer 1

x - y^2 = 0 ---> x = y^2 ( parabola opening to the right, vertex at  (0,0)) ;  let point on parabola that minimizes distance from (0,3) be (a,b); then, slope of line joining (0,3) and (a,b) = m =  -(3 - b)/a = (b - 3)/a;  using a = b^2, we have the following:  m = (b -3)/(b^2);  next, slope of line tangent to parabola at (a,b) is given by (dy/dx) at (a,b);  applying derivative operator, d/dx to x = y^2 ---> 1 = 2y(dy/dx) ---> dy/dx = 1/(2y) ---> dy/dx at (a,b) = M = 1/(2b);

but mM = -1 (lines mutually perpendicular) --->

[(b - 3)/(b^2)][1/(2b)] = -1 ---> (b - 3) = -2b^3 --->

2b^3 + b - 3 = 0;

try b = 1 ---> this works ---> a = b^2 ---> a = 1

distance in question = D = sqrt[(a - 0)^2 + (3 - b)^2] --->

D = sqrt[(1 - 0)^2 + (3 - 1)^2] = sqrt(1 + 4) = sqrt(5);  q.e.d.