determine the points on y=x2+1 that ARE CLOSEST TO(0,2)

Have a look at this graph, Melody.

https://www.desmos.com/calculator/blatwtgdae

I have graphed the tangent line to the curve at (1/√2, 3/2)  and a line perpendicular to this one that goes through the same point. And the perpendicular line also goes through (0,2). And the shortest distance between a line and a point is a perpendicular line to the given line passing through that point....

{The same argument can be made for the other point (-1/√2, 3/2)...}

Call the point(s) on the parabola that are closest to (0,2), (x, x^2 + 1)

And the distance between these two is given by.....

d = [(x-0)^2 + (x^2 + 1 - 2)^2 ]^(1/2)

d = [(x^2  + (x^2 -1)^2]^(1/2)

d = [(x^2 + x^4 - 2x^2 + 1]^(1/2)

d = [x^4 - x^2 + 1]^(1/2)

We wish to minimize this.....take the derivative....we have

d' = (1/2)[x^4 - x^2 + 1]^(-1/2)*(4x^3 - 2x)    set to 0

This will be 0 when the numerator= 0    so we have

(4x^3 - 2x) = 0    factor

2x(2x^2 -1 ) = 0   set each factor to 0

So, either x = 0     or x = ±√(1/2)

And when x = 0   y =(0)^2 + 1  = 1   so (0,1) is one possible point

And when x = ±√(1/2), y =  (±√(1/2))^2 + 1   = 3/2

So the other two possible points are (-√(1/2), 3/2) and (√(1/2), 3/2)

Check the distance between(0, 2) and (0, 1) =1

Check the distance between (0,2) and (√(1/2), 3/2)

d = [(√(1/2))^2 + (2 -3/2)^2]^.5  = [(1/2) + (1/2)^2]^.5 = [(1/2 + 1/4]^.5 = (3/4)^.5 = √3/2 ...    and this distance will be the same  between  (0,2) and (-√(1/2), 3/2)

And these two distances are less than the distance between (0,2) and (0,1)

So, the two points on the parabola y = x^2 + 1 that are closest to (0,2) are (±√(1/2), 3/2)

Thanks Chris,

I'm all confused, algegraically what you say seems correct but look at the pic.

I think the answer is  (0,0), (1,2) and (-1,2)My algebra is slightly diff from your but I get the same answer - i am sure it is wrong. 

It's doing my head in - I am going to take a break.  :))

https://www.desmos.com/calculator/ykxhu3rflb