Hi, I'm trying to solve the following problem:
Find the point on the parabola that is closest to the point (3, 0).
Thanks for your help.
D^2 = (x-3)^2 + y^2Now replace y by x^2, and expand to get:
D^2 = x^4 + x^2 - 6x + 9Differentiate and put the result equal to zero:
4x^3 + 2x - 6 = 0which has a solution x = 1.
The second derivative is positive at this point, giving a minimum value of D^2 (and hence D) at the point (1, 1).
Since by the definition of our equation and the fact that the point can be represented as we can replace the y with the x^2 as in the following:
Then we differentiate and set the derivitive equal to zero to find posibile soultions (i.e. the critical points). This is done here:
First note that:
And the derivitive is:
I think this derivitive is diffirent than what the previous poster put before me, but this is because he took a different approach, the answer will be the same either way. Anyway, the zero of this function is 1. Thus, the distance is shortest at x=1, and this distance is the following:
And thats the answer, I do believe.