Thread: Distance d from P

Let P = (x, y) be a point on the graph of y = x^2 + 2.

Express the distance d from P to the point (2, 0) as a function of x.

Originally Posted by symmetry

Let P = (x, y) be a point on the graph of y = x^2 + 2.

Express the distance d from P to the point (2, 0) as a function of x.

Distance between two points (x1y1) and (x2,y2):
d = sqrt[(x2 -x1)^2 +(y2 -y1)^2] ---------------------------------(i)

or
d = sqrt[(difference of the x's)^2 +(difference of the y's)^2] ------(ii)
Meaning, (x1 -x2)^2 is also (x2 -x1)^2; (y2 -y1)^2 is same as (y1 -y2)^2.

The point P(x,y) on the graph of y = x^2 +2
is P(x,(x^2 +2)) in terms of x.

The other point is (2,0)

So,
d = sqrt[(x -2)^2 +((x^2 +2) -0)
d = sqrt[(x-2)^2 +(x^2 +2)]
d = sqrt[(x^2 -4x +4) +(x^2 +2)]
d = sqrt[x^2 -4x +4 +x^2 +2]
d = sqrt[2x^2 -4x +6] --------------------answer.

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Edit.

Oooppps, wrong. I forgot to square the difference of the y's.

It should have been like this:
d = sqrt[(x -2)^2 +((x^2 +2) -0)^2]
d = sqrt[(x-2)^2 +(x^2 +2)^2]
d = sqrt[(x^2 -4x +4) +(x^4 +4x^2 +4)]
d = sqrt[x^2 -4x +4 +x^4 +4x^2 +4]
d = sqrt[x^4 +5x^2 -4x +8] --------------------final answer.

I thank you for every for the insight and math tips.

Basically, we apply the distance formula.

Good job!

Thank!