Find the point on the parabola y^2=x that is closest to the point (1,7?
let the point of contact be P(a,b)
let A be (1,7)
slope of AP = (b-7)/(a-1)
the closest point would be where the tangent at P is perpendicular to AP
2y dy/dx = 1
dy/dx = 1/(2y)
so at P, the slope of the tangent is 1/(2b)
so (b-7)/(a-1) = -2b
b-7 = -2ab + 2b
-b = -2ab + 7
2ab - b = 7
b(2a - 1) = 7
b = 7/(2a-1) **
but since (a,b) lies on y^2 = x
b^2 = a
subbing that into **
b = 7(2b^2 - 1)
2b^3 - b = 7
2b^3 - b - 7 = 0
nasty:
Wolfram said, b = 1.6279
then a = 2.6501
http://www.wolframalpha.com/input/?i=solve+b%5E3+-+b+-+7+%3D+0
or
let the distance be D, and (x,y) as the closest point
D^2 = (x-1)^2 + (y-7)^2
but x=y^2
D^2 = (y^2 - 1)^2 + (y-7)^2
2D dD/dy = 2(y^2 - 1)(2y) + 2(y-7)
= 4y^3 - 4y + 2y - 14
= 4y^3 - 2y - 14
dD/dy = (2y^3 - y - 7)/D
= 0 for a min of D
2y^3 - y - 7 = 0
notice I got the "same" equation, except my y was b, and my x was a
the closest point is (2.6501, 1.6279)
To find the point on the parabola y^2 = x that is closest to the point (1,7), we first need to understand some basic principles.
The distance between two points in the plane can be found using the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, we want to find the minimum distance between the point (1,7) and a point on the parabola y^2 = x. Let's represent the unknown x-coordinate of the point on the parabola as "a." Now we have the coordinates: (a, sqrt(a)).
The distance between these two points is given by:
Distance = sqrt((a - 1)^2 + (sqrt(a) - 7)^2)
Our goal is to find the value of "a" that minimizes this distance. To do that, we can differentiate the distance equation with respect to "a" and set the derivative equal to zero. Let's calculate it step by step:
1. Expand and simplify the distance equation:
Distance = sqrt((a - 1)^2 + (sqrt(a) - 7)^2)
Distance^2 = (a - 1)^2 + (sqrt(a) - 7)^2
2. Differentiate both sides of the equation with respect to "a":
2 * Distance * (d(Distance)/da) = 2 * (a - 1) + 2 * (sqrt(a) - 7) * (1/(2 * sqrt(a)))
Cancel out common terms:
Distance * (d(Distance)/da) = (a - 1) + (sqrt(a) - 7) * (1/(sqrt(a)))
3. Set the derivative equal to zero:
(a - 1) + (sqrt(a) - 7) * (1/(sqrt(a))) = 0
Simplify the equation:
a - 1 + (sqrt(a) - 7)/(sqrt(a)) = 0
a - 1 + (1 - 7/sqrt(a)) = 0
Combine like terms:
a - 8/sqrt(a) = 0
4. Solve the equation for "a":
Multiply both sides by sqrt(a) to eliminate the square root:
a * sqrt(a) - 8 = 0
a * sqrt(a) = 8
Square both sides:
a^3 = 64
Take the cube root of both sides:
a = 4
Now that we have found the value of "a" as 4, we can substitute it back into y^2 = x to find the corresponding y-coordinate:
y^2 = 4
y = ±2
Therefore, the point on the parabola y^2 = x closest to the point (1,7) is (4, ±2).