Find Shortest Distance of y=x^2 from (4,0)

In summary, the shortest distance between (4,0) and y= x2 will be along the line that is perpendicular to the graph. However, x= 4 is for the point (4, 0), not any point on the graph! To solve for x0, take the derivative of the distance equation and set it equal to zero, which leads to a difficult equation for x0.
  • #1
rhey
6
0
shortest distance??

Homework Statement



find the shortest distance of y=x^2 from (4,0)

Homework Equations





The Attempt at a Solution



if y=x^2, then
y'=2x

where x=4
the slope is 8

the solution is

y=8(x-4)
if that is the tangent line \, the normal line would be

y=(1/8)(x-4)

after that.. I'm stuck!
 
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  • #2
One way is to write out the Euclidian distance from vector [x, x^2] to vector [4, 0], then minimize w/r/t x.
 
  • #3
EnumaElish, what rhey is trying to do is, I believe, more fundamental and a "nicer" method.

The shortest distance from (4,0) to y= x2 will be along the line that is perpendicular to the graph. HOWEVER, rhey, x= 4 is for the point (4, 0), not any point on the graph! Let [itex](x_0, x_0^2)[/itex] be the point on y= x2 closest to (4, 0). The derivative there, and the slope of the tangent line, is 2x0 so the slope of the normal line is [itex]-1/(2x_0)[/itex]. The equation of the line through (4, 0), perpendicular to y= x2 at [itex](x_0,x_0^2)[/itex] is [itex]y= -1/(2x_0)(x- 4)[/itex]. If that is to pass through [itex](x_0, x_0^2)[/itex], you must have y= [itex]x_0^2= -1/(2x_0)(x_0- 4)[/itex]. Solve that equation for x0 and you are almost done!

Hmm, that leads to a rather difficult equation for x0! Darn it, EnumaElish may be right!
 
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  • #4
i really can't understand what are u trying to tell me.. honestly, I'm not that good in math! but I'm trying..
 
  • #5
1) Setup an equation for distance between point (0,4) and the function, [tex]f(x) = x^2[/tex]

[tex]d = \sqrt{(x_2 - x_1)^2 - (y_2 - y_1)^2}[/tex]
[tex]d = \sqrt{(x_2 - 4)^2 - (y_2)^2}[/tex]
[tex]d = \sqrt{(x_2 - 4)^2 - (x_2)^4}[/tex]

2) Take the derivative of d and set it equal to zero

[tex]d' = 0 = \frac{4(x_2)^3 - 2(x_2 - 4)}{2\sqrt{(x_2 - 4)^2 - (x_2)^4}}[/tex]
[tex]0 = 4(x_2)^3 - 2(x_2 - 4)[/tex]
[tex]4(x_2)^3 = 2(x_2 - 4)[/tex]
[tex]{x_2}^3 = x_2 - 4[/tex]

As others have said, the problem is still difficult to solve from here...
 
Last edited:
  • #6
And, as EnumaElish said, it's far easier to minimize the square of the distance, not the distance itself! Oh, and it should be y- 4, not x- 4.
 

1. How do you find the shortest distance of y=x^2 from (4,0)?

To find the shortest distance, you can use the formula d = |(ax0 + by0 + c)|/√(a^2 + b^2) where (x0, y0) is the point (4,0) and the line is y=x^2. This formula is derived from the distance between a point and a line in coordinate geometry.

2. What is the equation of the line y=x^2?

The equation of the line y=x^2 is a quadratic function in the form of y=ax^2+bx+c, where a=1, b=0, and c=0. This means that the line is a parabola with its vertex at the origin and opens upwards.

3. How do you determine the coordinates of the point of intersection between y=x^2 and the line perpendicular to it passing through (4,0)?

To find the point of intersection, you can solve the system of equations formed by y=x^2 and the perpendicular line. This can be done by substituting the value of x from the perpendicular line equation into the equation of y=x^2. The resulting quadratic equation can be solved to find the x-coordinate of the point of intersection, and the y-coordinate can be found by substituting the x-value into either of the original equations.

4. Can the shortest distance of y=x^2 from (4,0) be negative?

No, the shortest distance cannot be negative. Distance is a positive quantity and cannot have a negative value. The formula d = |(ax0 + by0 + c)|/√(a^2 + b^2) ensures that the distance is always positive.

5. Is there a shortcut to finding the shortest distance of y=x^2 from (4,0)?

Yes, there is a shortcut called the perpendicular distance formula. This formula states that the shortest distance from a point (x0, y0) to a line y=mx+c is given by d = |y0-mx0-c|/√(m^2+1). This formula can be applied directly to find the shortest distance of y=x^2 from (4,0) without having to derive the formula each time.

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