Thread: Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

We want to minimize the distance between and . We know that





To make things easier, . To find the x that minimizes , you may as well find the x that minimizes . So find the (x,y) coordinate that minimizes .

Originally Posted by Roleparadise

Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

Let required point be (x,y) Let D= distance from (1,5) So D^2=(x-1)^2+(y-5)^2 But (x,y) is on curve so y=x^2+3 giving D^2=(x-1)^2+(x^2-2)^2
Want D^2 to be minimum (then D will be minimum) So want derivative=0
Hence want 2(x-1)+2(x^2-2)2x=0 2x-2+4x^3-8x=0 4x^3-6x-2=0 Using a numerical change of sign method I get x=1,366 so y=4.866

I dont agree with your expression for D

Whoops, I did instead of ...

Anyway, . Take derivative with respect to x and set to zero:

.

There are three real solutions for x: x = -1, -.366, and 1.366. Check each (x,y) point to see which one is closest.

Originally Posted by Roleparadise

Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

Here is an entirely different way to solve it
Find a point on the curve where the tangent is perpendicular to the normal through .

Thus solve .