find the point on the parabola y=x^2 nearest to the point (2,-8)

Question

shamil98 · Answer

Graph it, and look at the graph?

anonymous · Answer

Hint: The point on the curve which is closest to the given point will have the property that the line drawn between these two points will be perpendicular to the tangent to the curve at that point.

anonymous · Answer

Thus, if (x1,y1) on the parabola is closest to (2,-8) then the line from (x1,y1) to (2,-8) will be perpendicular to the tangent to the parabola at (x1,y1).

Also, if two lines are perpendicular, then the product of their slopes is -1.

I think with this we have enough information to solve this problem.

anonymous · Answer

Here are the details:

|dw:1392533734676:dw|

We then have to solve of x1 in this cubic equation, using a calculator [perhaps].

whpalmer4 · Answer

Another approach, if you know some calculus, is to use the distance formula to make a function for the distance of a point on the parabola to $(2,-8)$.  Take the first derivative and set it equal to 0 to find the value of $x$ such that the distance is minimized.  Same result.

anonymous · Answer

@whpalmer : Yes an in both cases you need to finally solve the SAME cubic equation - something I did not expect!