HELP PLEASE
Find the minimum distance from the point (4, 2) to the parabola y^2=8x
eq. 1 - - - > (y^2)=8x - - - > x=(y^2)/8
eq. 2 - - - -> point distance formula
d=[(x2-x1)^2+(y2-y1)^2] ^1/2
input (4,2) and eq. 1 to eq. 2
d=[((y^2)/8 - 4)^2 + (y - 2)^2] ^1/2
differentiate d
let u = ((y^2)/8 - 4)^2 + (y - 2)^2
du = 2((y^2)/8 - 4)^(2-1) ((2y) /8)
+ 2(y - 2)^(2-1) (1)
d= (u) ^1/2
d'=(1/2) (u) ^(1/2 - 1) (du)
d'=du/[2(u)^(1/2)]
then rule of minima maxima equate d' to 0
0 = du/[2(u)^(1/2)]
cross multiplication
0=du
0=2((y^2)/8 - 4)((2y)/8) + 2(y-2)
0=(4y^3)/64 - 2y + 2y - 4
y^3= 64 - - - > y=4
input 4 to eq. 2
answer is 2(2)^1/2
2sqrt2
To find the minimum distance from a point to a curve, we can use the concept of perpendicular distance. We need to find the point on the parabola that is closest to the given point.
Here are the steps to find the minimum distance:
Step 1: Convert the equation of the parabola into a function of x so that we can express y in terms of x.
Given equation: y^2 = 8x
Rearrange the equation: x = y^2/8
Step 2: Calculate the derivative of this function because the slope of the line connecting the closest point on the parabola to the given point should be perpendicular to the tangent of the parabola.
Differentiate x = y^2/8 with respect to y:
dx/dy = (1/8) * 2y = y/4
Step 3: Calculate the slope of the line connecting the given point (4, 2) to any point (x, y) on the parabola.
Slope of the line = (y - 2)/(x - 4)
Step 4: Set the two slopes equal to each other and solve for y.
y/4 = (y - 2)/(x - 4)
Cross multiply: (x - 4)y = 4(y - 2)
Simplify: xy - 4y = 4y - 8
Rearrange: xy = 8
Step 5: Substitute the value of xy from the previous equation into the equation of the parabola.
8 = y^2/8
Rearrange: y^2 = 64
Step 6: Solve for y by taking the square root of both sides.
y = ±√64 = ±8
Step 7: Substitute the values of y into the equation of the parabola to find the corresponding x-coordinates.
When y = 8, 8^2 = 8x gives x = 8
When y = -8, (-8)^2 = 8x gives x = 8
So, we have two points (8, 8) and (8, -8) on the parabola that are equidistant from the given point (4, 2).
Step 8: Calculate the distance between the given point and each of the obtained points using the distance formula.
Distance between two points (x1, y1) and (x2, y2) = √[(x2 - x1)^2 + (y2 - y1)^2]
Distance1 = √[(8 - 4)^2 + (8 - 2)^2] = √[(4)^2 + (6)^2] = √[16 + 36] = √52
Distance2 = √[(8 - 4)^2 + (-8 - 2)^2] = √[(4)^2 + (-10)^2] = √[16 + 100] = √116
Step 9: Compare the distances and determine the minimum distance.
The minimum distance is the smaller of the two distances: √52
Therefore, the minimum distance from the point (4, 2) to the parabola y^2 = 8x is √52.
(y^3)/4 -4 =0
isnt that supposed 2 be
(y^3)/16 -4 = 0
Yes. So y^3 = 64 and y = + or - 4
Nice job catching that mistake!
x = 2, either way.
One of the two points may be a relative minimum only
2sqrt2
The square of the distance from the point to the line is
D^2 = (x-4)^2 + (y-2)^2
Let y and x(y) vary and minimize that.
D^2 = [(y^2/8)-4]^2 +(y-2)^2
D^2 = (y^4)/64 -y^2 + 16 +y^2 -4y +4
= y^4/64 - 4y +20
d(D^2)/dy = 0
(y^3)/4 -4 =0
y^3 = 16
y = 2*2^(1/3) = 2.5196
x = 0.79355
Verify (or correct) my work and compute D^2 , and then D