Help on finding the point(s) on the parabola y=2x^2 closest to (0,1)?

Question

And how do I find the distance between these points.
Thanks
Dave

Answer 1

D = distance between (0,1) and (p,2p^2) on the curve

D^2 = p^2 + (2p^2 - 1)^2
d(D^2)/dp = 2p + 2*4p*(2p^2 - 1)
= 2p + 8p*(2p^2 - 1)

This is 0 when p = 0
and when p^2 = 3/8

When p = 0, D^2 = 1
When p = sqrt(3/8), D^2 = 0.4375

So the minumum distance involves the point
[+/- sqrt(3/8), 3/4]
the distance is sqrt(7)/4

Answer 2

Before we begin, recognize that the parabola in question is symmetrical along the line x=0.
Because the right half of the parabola is identical to the left half of the parabola, there will be two points that are closest to (0,1), one on the left side and one on the right side.
So if we find that the point (a,b) is closest to (0,1), the point (-a,b) will also be equally far away.

Say you have a point (a,b) that is on the parabola: y=2*(x^2)
According to the parabola equation, whatever value x is, means y is 2*(x^2)
In our case, a is x, and b is y.

Therefore, b=2*(a^2)

We can therefore write that this point (a,b) must have coordinates:
( a , 2*(a^2) )

Notice that for every value of x, there is exactly one corresponding point on the parabola. Also notice that every point along the parabola has exactly one corresponding value for x. In our case, the x-value is the variable [a]. So we can try to find the value of [a] that corresponds to the point closest to (0,1).

The distance between two points can be determined with the Pythagorean theorem. Imagine a right angled triangle that has it's diagonal (hypotenuse) connecting the two points.

In this case, our two points are:
( a , 2*(a^2) )
and
( 0 , 1 )

The horizontal distance (delta_x) between the two points is:
delta_x = a - 0
which simplifies to:
delta_x = a

The vertical distance (delta_y) between the two points is:
delta_y = 2*(a^2) - 1

Using the Pythagorean theorem,
diagonal^2 = delta_x^2 + delta_y^2

Now, the diagonal of our triangle is exactly the line connecting the two points, so diagonal=distance. This leads to:
distance^2 = delta_x^2 + delta_y^2

Recall that we can calculate delta_x and delta_y based on the value of the variable a:
distance^2 = a^2 + ( 2*(a^2) - 1 )^2

Now we expand:
distance^2 = a^2 + 4*(a^4) - 4*(a^2) + 1

Collect terms:
distance^2 = 4*(a^4) - 4*(a^2) + a^2 + 1
distance^2 = 4*(a^4) - 3*(a^2) + 1

Let's review what we're trying to do here. We know that for every value of the variable a, there is a corresponding point at:
( a , 2*(a^2) )

We know that the distance^2 between that point and the (0,1) point is:
distance^2 = 4*(a^4) - 3*(a^2) + 1

If we can find the value for the variable a that makes the smallest value for distance, we're done.

To do this, we need to make one more observation. Imagine we find the point with the smallest distance^2.
Say:
distance^2 = z

This point must also be the point with the smallest distance,
distance = sqrt{z}

Since there is no point with a distance smaller than sqrt{z}, there cannot also be a distance^2 smaller than z.
The conclusion here is that finding the smallest distance^2 gives the same point as finding the smallest distance.

So let's begin the final steps.

distance^2 = 4*(a^4) - 3*(a^2) + 1

We notice that this looks kind of like a parabola, but not quite. We're going to do a mathematical trick here:

Say we let: u=a^2
Each value of (u) corresponds to two values of (a)
a = -sqrt{u}
and
a = +sqrt{u}

Each of these values corresponds to two points:
( +sqrt{u} , 2*u )
and
( -sqrt{u} , 2*u )

So for each value of u, there are two points on the parabola. But remember, because of the symmetry, both points are the SAME DISTANCE from the point (0,1) after all!

This means, for each value of u, we are given two points each with the same distance from the point (0,1). This distance is:
distance^2 = 4u^2 - 3u + 1

If we can find the value for the variable u that gives the least distance^2, we will have BOTH points with the least distance. Since we want to find the minimum distance^2, we simply need to find the minumum value of:
4u^2 - 3u + 1

The above expression is in fact also a parabola, so we find the minimum by completing the square.
distance^2 = 4u^2 - 3u + 1

First, factor the 4:
distance^2 = 4*( u^2 - 0.75u + 0.25 )

Next, complete the square on the inside:
distance^2 = 4*( (u-0.375)^2 + 0.25 - 0.375^2 )
distance^2 = 4*( (u-0.375)^2 + 0.25 - 0.140625 )
distance^2 = 4*( (u-0.375)^2 + 0.109375 )

Then, multiply the 4 back in
distance^2 = 4*(u-0.375)^2 + 0.4375

Let's call it:
distance^2 = parabola_u

Square root both sides:
distance = sqrt{ parabola_u }

So we're done. If we know the minimum value of parabola_u, we know the minimum distance. If we know the corresponding value of u, we can figure out the points that are closest.

1. What is the minimum value of parabola_u?
parabola_u = 4*(u-0.375)^2 + 0.4375
So, the minimum value is 0.4375.
This means the minimum distance is sqrt{ minimum_value } which works out to be about 0.66143783...

2. Where is this point exactly?
Well, in order for parabola_u to be at its minimum value, u must be 0.375. Our two points corresponding to each u value are:
( +sqrt{u} , 2*u )
and
( -sqrt{u} , 2*u )
This means the two closest points are:
( +sqrt{0.375} , 2*0.375 )
and
( -sqrt{0.375} , 2*0.375 )

which works out to be about,
( 0.61237244 , 0.75 )
and
( -0.61237244 , 0.75 )


-------------
Recap
-------------

(a,b) is on the parabola

b=2*(a^2)

the point is,
( a , 2*(a^2) )

The distance^2 between that point and the (0,1) point is:
distance^2 = 4*(a^4) - 3*(a^2) + 1

Let u=a^2, then the points are,
( +sqrt{u} , 2*u )
and
( -sqrt{u} , 2*u )

NOTE: These two points are equally far away from (0,1)

Using the u variable instead of the a variable is handy, because:
distance^2 = 4u^2 - 3u + 1 = parabola_u

So if,
parabola_u = 4u^2 - 3u + 1
we can complete the square:
parabola_u = 4*(u-0.375)^2 + 0.4375

This implies that parabola_u is smallest at u=0.375, and the smallest value of parabola_u is 0.4375
And distance^2 = parabola_u, which implies that distance^2 is smallest at u=0.375.
The smallest distance^2 is 0.4375.

And when distance^2 is smallest, distance is also smallest, so the smallest distance is sqrt{0.4375}, which works out to be about 0.66.
And when u=0.375,
a = -sqrt{0.375}
and
a = +sqrt{0.375}

Which means the closest points are:
( +sqrt{0.375} , 0.75 )
and
( -sqrt{0.375} , 0.75 )

which works out to be about
( 0.61 , 0.75 )
and
( -0.61 , 0.75 )



-------------
Answer
-------------

The points on the parabola closest to (0,1) are exactly:
( +sqrt{0.375} , 0.75 )
and
( -sqrt{0.375} , 0.75 )

They have a distance of sqrt{0.4375}.


-------------
Note
-------------
This question is simpler to solve using calculus. You can avoid the u=a^2 step. Take the derivative of distance^2 to find its critical points, and pick the minimums.

Answer 3

Find the minimum distance between the point P(0, 1) and the parabola y = 2x². What are the closest points on the parabola to P?

Since the given point P(0, 1) is on the line of symmetry of the parabola, there will be two closest points.

Let
D = distance
P(0, 1)
Q(x, y) = Q(x, 2x²) = point on curve

We have:

D² = (x - 0)² + (2x² - 1)² = x² + 4x^4 - 4x² + 1
D² = 4x^4 - 3x² + 1

Take the derivative and set it equal to zero to find the critical points.

2D(dD/dx) = 16x³ - 6x
dD/dx = (16x³ - 6x)/(2D) = (8x³ - 3x)/D = 0
8x³ - 3x = 0
x(8x² - 3) = 0
x = 0, ±√(3/8)

The second derivative determines the nature of the critical points.

dD/dx (8x³ - 3x)/D
d²D/dx² = (24x² - 3)/D

For x = 0
d²D/dx² = -3/D < 0
This is a relative maximum. This is not what we want. The solution is rejected.

For x = ±√(3/8)
d²D/dx² = [24(3/8) - 3]/D = 6/D > 0
This is a relative minimum which is what we want.

So the solution is:
x = 0, ±√(3/8)

y = 2x² = 2(3/8) = 6/8 = 3/4

The points of closest approach on the curve are:
[√(3/8), 3/4] and [-√(3/8), 3/4].

The minimum distance is the same in both cases. It is:

D = √[(√(3/8) - 0)² + (3/4 - 1)²] = √(3/8 + 1/16) = √(7/16)
D = √7 / 4

Answer 4

Draw the oblong with coordinates D = (0,0),C = (3,0), B = (3,3), and A = (0,3). If the coordinates of P are (x, y), then the coordinates of Q would desire to be (3-x, 3-y). So PQ = DP turns into 9 - 12x + 4x² + 9 - 12y + 4y² = x² + y² 3x² - 12x + 3y² - 12y = -18 x² - 4x + y² - 4y = -6 (x-2)² + (y-2)² = 2 So P would desire to lie on the arc of the circle with midsection at (2,2) and radius ?2. The furthest left that factor P is additionally is a factor tangent to the circle (x-2)² + (y-2)² = 2 at a factor on the brink of A. enable O = (2, 2) be the midsection of that circle. Then DO = 2?2, OP = ?2 and triangle DOP is a ideal triangle. Triangle DOP is for this reason a 30-60-ninety ultimate triangle and the degree of perspective PDO is 30°. because of the fact the degree of perspective ODC is 40 5°, it follows that the degree of perspective ADP is 15°

Answer 5

Isn't the closest point the vertex (0,0) The distance would be 1 unit. Graph the two points vertex(0,0) and (0,1) and you will see the distance is 1 unit. It's a parabola facing upwards.

Answer 6

1. Draw the parabola...easy to do.

2. To find the points on the parabola that are closest to (0,1), solve 1 = 2x^2. The two points that are closest to as well as equidistant from (0,1) are (0.7,1) and (-0.7,1).

3. Use distance formula.

distance = ((y2-y1)^2 + (x2+x1)^2)^0.5

If you get distance = 0.7 units, you are right.

I hope this helps.

Answer 7

oopsy. i saw my huge flaw.