Find the point on the curve y^2=4x closest to the point (8,0).
My Work: Using the Distance formula-
my answer i got was (7.821,5.593) but idk if i did it right...
3 answers
one way to tell whether you did it right, is to show your work! Then we can see how you arrived at your result.
Ok so i set y=(4x)^1/2
Using the distance formula...
i got it down to d=((x-8)^2+((4x)^2)^1/2
Took the derivative and got dy/dx=2x+1/((x)^1/2)-16
Using the distance formula...
i got it down to d=((x-8)^2+((4x)^2)^1/2
Took the derivative and got dy/dx=2x+1/((x)^1/2)-16
hmmm. better see my solution to your previous posting.
the distance should have (x-8)^2 + 4x, not (4x)^2, since 4x is already y^2!
the distance should have (x-8)^2 + 4x, not (4x)^2, since 4x is already y^2!