Find the point on the curve y^2=4x closest to the point (8,0).
My Work: Using the Distance formula-
my answer i got was (7.821,5.593) but idk if i did it right...
One way is to find the length of the line normal to the curve which passes through the point.
The tangent at any point has slope 1/√x, so the normal line has slope -√x.
The line through (8,0) and (x,2√x) has slope 2√x/(x-8), which we want to be -√x.
2√x/(x-8) = -√x
-2 = x-8
x = 6
The point is (6,2√6)
Or, using the distance formula,
d = √((x-8)^2+(y)^2)
= √(((x-8)^2+4x)
= √(x^2-12x+64)
d' = (x-6)/√(x^2-20x+64)
minimum distance d is where
d'=0 at x=6, as above
Ignore the comment about the length of the normal line. The question did not actually ask for the distance, just the point on the curve.
We just wanted to find the point where the normal to the curve passes through (8,0)
Oh so then i plug x=6 back to the original to get Y
To find the point on the curve y^2 = 4x closest to the point (8, 0), you need to use the distance formula. Your answer seems close, but let's go through the steps together to make sure.
Step 1: Start with the equation of the curve: y^2 = 4x
Step 2: Identify the coordinates of the given point: (8, 0)
Step 3: Let's assume the point on the curve closest to (8, 0) is (a, b). This means we need to find the values of 'a' and 'b'.
Step 4: Apply the distance formula. The distance between the points (8, 0) and (a, b) is given by the formula:
Distance = √((a - 8)^2 + (b - 0)^2)
Step 5: Since we want the point on the curve, substitute the value of 'b' using equation (1):
√((a - 8)^2 + (sqrt(4a))^2)
Step 6: Simplify the equation:
Distance = √((a - 8)^2 + 4a)
Step 7: To minimize the distance, we differentiate the equation with respect to 'a'. Let's find the derivative:
d(Distance)/da = d(√((a - 8)^2 + 4a))/da
Using the chain rule and simplifying, we get:
d(Distance)/da = (2a - 8) / (2√((a - 8)^2 + 4a))
Step 8: Set the derivative equal to zero and solve for 'a':
(2a - 8) / (2√((a - 8)^2 + 4a)) = 0
2a - 8 = 0
2a = 8
a = 4
Step 9: Substitute the value of 'a' back into equation (1) to find 'b':
b^2 = 4(4)
b^2 = 16
b = ±4
So, we have two potential points: (4, 4) and (4, -4).
Step 10: Calculate the distance between each potential point and the given point (8, 0) using the distance formula:
Distance1 = √((4 - 8)^2 + (4 - 0)^2)
Distance1 ≈ 7.211
Distance2 = √((4 - 8)^2 + (-4 - 0)^2)
Distance2 ≈ 7.211
Step 11: Compare the distances to determine which one is closer. In this case, both distances are the same (approximately 7.211), so both points are equidistant.
Therefore, the two points (4, 4) and (4, -4) on the curve y^2 = 4x are equidistant to the point (8, 0).