To find the nearest point on the curve we're looking for the line that is perpendicular to the tangent (normal). First we find the gradient by differentiation: 2x.
The normal has a gradient of -1/(2x) because the product of the slope of the normal and the slope of the tangent (gradient) is -1 and 2x*(-1/(2x))=-1. The gradient of the normal varies according to the position on the curve.
The equation of the normal must pass through the point A. So y=mx+b must be satisfied by x=0 and y=3: so b=3 and the equation is y=mx+3, where m is to be determined. This line meets the curve when x^2=mx+3. If we put m=-1/(2x), we get x^2=-1/2+3=5/2 and x=+sqrt(2.5) and y=2.5. Therefore the nearest point is either (sqrt(2.5),2.5) or (-sqrt(2.5),2,5). Since the parabola is symmetrical the points are equidistant from A because A is on the axis of symmetry. The normal at (-sqrt(2.5),2.5) is the reflection of the normal at (sqrt(2.5),2.5). [The equations of the two normals are y=3-x/sqrt(10) (right-hand) and y=3+x/sqrt(10) (left-hand). The distance from A=sqrt(2.75).]