How do you find the points on the parabola #y = 6 - x^2# that are closest to the point (0,3)?

Question

Cameron Alexander · Accepted Answer

Substitute the equation of the parabola into the distance formula to get the square root of a quartic to minimise. This is quadratic in #x^2#, so complete the square to find the minimum.

If #(x, y)# is a point on the parabola, then the distance between #(x, y)# and #(0, 3)# is: #sqrt((x-0)^2+(y-3)^2)# #= sqrt(x^2+(6-x^2-3)^2)# #=sqrt(x^2+(3-x^2)^2)# #=sqrt(x^2+9-6x^2+x^4)# #=sqrt(x^4-5x^2+9)# #=sqrt((x^2-5/2)^2+11/4)# This will have its minimum value when #(x^2-5/2)^2 = 0#, that is when #x = +-sqrt(5/2) = +-sqrt(10)/2# When #x = +-sqrt(10)/2# we have #y = 6 - x^2 = 6 - 5/2 = 7/2# So the points on the parabola at minimum distance from #(0, 3)# are: #(+-sqrt(10)/2, 7/2)# As a bonus, we have also calculated the minimum distance as: #sqrt(11/4) = sqrt(11)/2#

Emilia Aguilar · Answer

undefined To find the points on the parabola $ y = 6 - x^2 $ that are closest to the point $ (0,3) $, you need to minimize the distance between the given point and any point on the parabola. This can be achieved by minimizing the square of the distance between the points. Therefore, you need to minimize the square of the distance function $ d(x) = (x - 0)^2 + (6 - x^2 - 3)^2 $. To do this, find the critical points of $ d(x) $ by setting its derivative equal to zero and then determine which critical points correspond to the minimum distance.