Answer:
(4, 4)
Step-by-step explanation:
There are a couple of ways to go at this:
- Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
- Find the equation of a normal line to the parabola that goes through the given point.
1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...
... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x
Differentiating with respect to x and setting dd/dx=0, we have ...
... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)
We can factor 2 from this to get
... 0 = x -2 +(y -8)(dy/dx)
Differentiating the parabola's equation, we find ...
... 2y(dy/dx) = 4
... dy/dx = 2/y
Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get
... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y
... 64 = y³ . . . . . . multiply by 4y, add 64
... 4 = y . . . . . . . . cube root
... y²/4 = 16/4 = x = 4
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2. The derivative above tells us the slope at point (x, y) on the parabola is ...
... dy/dx = 2/y
Then the slope of the normal line at that point is ...
... -1/(dy/dx) = -y/2
The normal line through the point (2, 8) will have equation (in point-slope form) ...
... y - 8 = (-y/2)(x -2)
Substituting for x using the equation of the parabola, we get
... y - 8 = (-y/2)(y²/4 -2)
Multiplying by 8 gives ...
... 8y -64 = -y³ +8y
... y³ = 64 . . . . subtract 8y, multiply by -1
... y = 4 . . . . . . cube root
... x = y²/4 = 4
The point on the parabola that is closest to the point (2, 8) is (4, 4).
tells us that the vertex has moved left 3 units.