Minimum distance from a point to a parabola?

Question

Can anyone please suggest steps on how to solve that kind of problem? Specifically, I'm solving for the minimum distance between the parabola y^2 = 8x and the point (6,0). Thanks! :-)

Answer 1

You want to minimize (x-6)^2+y^2, which is the squared distance
between your point and a point (x,y), knowing that y^2=8x. So you actually want to minimize (x-6)^2+8x as a function of x. You differentiate and get f'(x)= 2(x-6)+8 which vanishes for x=2. Hence your minimum distance is sqroot{16+16}= 4 sqroot{2}. O.K.?

Answer 2

So you want to solve for the minimum distance between the parabola

y^2 = 8x

And the point (6,0).

Recall that the distance formula between two points is given by

d = sqrt ( [x2 - x1]^2 + [y2 - y1]^2)

As in any min/max problem, we need to express one variable in terms of the other. Note that we have y^2 = 8x, which means
x = (1/8)y^2

Now, we plug in our values for the distance formula as normal. Remember that we want to find the distance between (6,0) and (x,y). The reason why I say (x,y) is because we want the *general* distance. So we plug in (6,0) for (x1,y1) and (x,y) for (x2,y2).

d = sqrt ( [x - 6]^2 + [y - 0]^2)
d = sqrt ( [x - 6]^2 + y^2)

HOWEVER, we can express x as x = (1/8)y^2; thus

d = sqrt ( [(1/8)y^2 - 6]^2 + y^2 )

So now that we have d expressed as a single variable, we can define this to be our function, d(y).

d(y) = sqrt ( [(1/8)y^2 - 6]^2 + y^2 )

Because this is perfectly valid, let's square both sides to get rid of the square root.

[d(y)]^2 = [(1/8)y^2 - 6]^2 + y^2

In order to find the minimum distance, we must obtain d'(y) and then make it 0. In this case, we have to differentiate implicitly.

2[d(y)] [d'(y)] = 2[(1/8)y^2 - 6] [(2/8)y] + 2y

Now, we make d'(y) = 0, effectively making everything on the left hand side 0.

0 = 2[(1/8)y^2 - 6] [(2/8)y] + 2y

Let's divide both sides by 2. Note that each term on the right hand side has a 2 in it, so they cancel out. Oh, and let's reduce that 2/8 while we're at it.

0 = [(1/8)y^2 - 6] [(1/4)y] + y

Now, let's distribute and expand.

0 = (1/32)y^3 - (6/4)y + y

Multiply both sides by 32 to get rid of all fractions, to get

0 = y^3 - 48y + 32y

Reducing,

0 = y^3 - 16
16 = y^3
Therefore, y = (16)^(1/3), or the cuberoot(16). This actually reduces to y = 2cbrt(2)

To actually *find* the minimum distance, all we have to do is plug y = 2cbrt(2) into our distance function.

d(y) = sqrt ( [(1/8)y^2 - 6]^2 + y^2 )
d(2cbrt(2)) = sqrt ( [(1/8)(2cbrt(2))^2 - 6]^2 + (2cbrt(2))^2 )

= sqrt ( [(1/2)(cbrt(4)) - 6)]^2 + 4cbrt(4) )

I'll leave it up to you to simplify this.

Answer 3

Do you know calculus?

One way is to determine the tangent, then the normal, then figure out when the normal goes through the point

Another way is:

distance from (6,0) to point (x,y) is sqrt((x-6)^2+y^2) but then since point (x,y) is on parabola we can substitute y^2=8x so then we need to minimize sqrt((x-6)^2+8x) and to make it easier we just minimize distance sqaured: minimize (x-6)^2+8x= x^2-4x+36 which has a minimum at x=2

Answer 4

3 feet