Question

find the point on the parabola 2y=x^2 that is closest to the point (-4,0)

Answer 1

a point on the curve looks like \((x, \frac{x^2}{2})\)

Answer 2

and the square of the distance between \((x, \frac{x^2}{2})\) and \((-4,0)\) is
\[d^2=(x+4)^2+(\frac{x^2}{2})^2\]
minimize that by taking the derivative, finding the critical points, etc

Answer 3

thank youu!

Answer 4

yw