Find the point on the parabole y=9-x^2 closest to the point (3,9)

(x,y)=(−1,1) is the closest point on y=x2 to (−3,0)

The distance from any point (x,y) to a point (ˆx,ˆy) is
XXX√(x−ˆx)2+(y−ˆy)2

For points on y=x2 this becomes
XXXd(x)=√(x−ˆx)2+(x2−ˆy)2
and
more specifically for the point (ˆx,ˆy)=(−3,0) this becomes
XXX√(x+3)2+(x2−0)2

XX=√x4+x2+6x+9

The problem is to minimize d(x)
or equivalently (but slightly simpler) to minimize
XXXf(x)=x4+x2+6x+9

The minimum occurs when f'(x)=0
That is when
XXX4x3+2x+6=0

An obvious (by inspection) root is x=−1
(and in fact there are no other real roots)

If x=−1
then y=x2=(−1)2=1

i hope you understand :)

stethere all not eveanp-by-step explanation:

hi! i think the answer is 1/2 : )