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Given point P moving along a curve y = x²-1 and point Q moving along a line y = x-3. Find the coordinate of point Q and distance between P and Q that makes the distance between two points have minimum value. Show your work, thank you!

User Dave De Jong
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1 Answer

23 votes
23 votes

Answer:

Part A

The coordinates of point Q that makes the distance between the two points have minimum value is (9/8, -15/8)

Part B

The distance between P and Q at the closest point is approximately 1.287 units

Explanation:

From the details given in the question, we have;

The curve along which point P is moving is, y = x² - 1

The (straight) line along which point Q is moving is, y = x - 3

The domain of the path of point P = All real numbers

The range of the path of point P = -1 < x ≤ ∞

The domain of the path of point Q = All real numbers

The range of the path of point Q = All real numbers

From the graph of the functions created with Microsoft Excel, we have that from point (0, 0) and moving to the positive x-axis, the slope of the path of P increases from 0 to ∞, while the slope of the path Q is constant and equal to 1

Given that the path of P is convex relative to the path of P, the two paths will be closest when they have equal slopes

Therefore at the closest point, for the path of point P, we have;

d(x² - 1)/dx = 1 (the slope of the path of point Q)

2·x = 1

∴ x = 1/2 on the path of point P at the closest point the path of point P to the path of point Q

∴ y = x² - 1 = (1/2)² - 1 = -3/4

y = -3/4

Given that two paths are parallel at the closest point, distance between the two paths is the perpendicular distance from the line y = x - 3 to the point (1/2, (-3/4)), which is given as follows;

∴ The slope of the line representing the perpendicular distance between the two paths = -1

Therefore, the equation of the line in point and slope form is given as follows;

y - (-3/4) = -1 × (x - 1/2)

y = -x + 1/2 - 3/4 = -x - 1/4

y = -x - 1/4

At the coordinates of point Q that makes the distance between the two points have minimum value, we have;

-x - 1/4 = x - 3

∴ 2·x = 3 - 3/4 = 2 1/4 = 9/4

x = 9/8

y = x - 3 = 9/8 - 3 = -15/8

Therefore, the coordinates of point Q that makes the distance between the two points have minimum value = (9/8, -15/8)

Part B

The formula for the length between two points is


l =\sqrt{\left (y_(2)-y_(1) \right )^(2)+\left (x_(2)-x_(1) \right )^(2)}

The distance, d, between P and Q at the closest point is given as follows;

d = √(((9/8) - (1/2))² + ((-15/8) - (-3/4))²) = √(53/32) ≈ 1.287

The distance between P and Q at the closest point, d ≈ 1.287

Given point P moving along a curve y = x²-1 and point Q moving along a line y = x-example-1
User Petrit
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