SOLUTION: Find the point of the parabola 9y = x^2 that is closest to the point (5,-2).
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Question 1184801: Find the point of the parabola 9y = x^2 that is closest to the point (5,-2).
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
the point of the parabola closest to the point (,)
simply, that will be intersection point of the parabola and vertical line that passes through the point (,)
=> (,) is the closest to the point (,)
Answer by greenestamps(13206) (Show Source): You can put this solution on YOUR website!
The solution from the other tutor is obviously incorrect -- as her graph shows. There are points on the parabola that are closer to (5,-2) than (5,25/9).
The fixed point is (5,-2); the variable point (on the curve) is (x,x^2/9).
We need to minimize the distance between the fixed point and the variable point. Since the distance formula involves square roots, we can minimize the square of the distance between the points.
The square of the distance between the two points is
Set the derivative equal to 0 to find the minimum
Solving this algebraically is difficult; a graphing calculator shows the minimum is when x=3. So the point on the curve closest to (5,-2) is (3,1).
ANSWER: (3,1) is the point on y=x^2/9 closest to (5,-2).
This can be verified by looking at slopes.
The derivative of the function at x=3 is 2x/9 = 6/9 = 2/3.
The slope of the line through (3,1) and (5,-2) is -3/2.
The two slopes are negative reciprocals, so the tangent to the curve and the line through (3,1) and (5,-2) are perpendicular, making (3,1) the point on the curve closest to (5,-2).
A graph, showing the tangent to the curve at (3,1) and the segment from (5,-2 to (3,1)....
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