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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.

User Psilocybin
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1 Answer

14 votes
14 votes

give data:

The currve moving along with point P is,


y=x^2-1

The line moving along the point Q is,


y=x-3

Let us now graph the given curve and line,

The closest point for the curve moving point P is,


\begin{gathered} (d(x^2-1))/(dx)=1 \\ 2x=1 \\ x=(1)/(2) \end{gathered}

thus,


\begin{gathered} y=x^2-1 \\ =((1)/(2))^2-1 \\ y=-(3)/(4) \end{gathered}

The distance between the two given paths is the perpendicular distance from the line y=x-3 to the point (1/2,-3/4).

The equation of the line in point and slope form is,


\begin{gathered} y-((3)/(4))=-1(x-(1)/(2)) \\ y=-x+(1)/(2)-(3)/(4)=-x-(1)/(4) \\ y=-x-(1)/(4) \end{gathered}

at the coordinates of point Q is the distance between the two points have minimum value.,


\begin{gathered} -x-(1)/(4)=x-3 \\ 2-x=3-(3)/(4)=2(1)/(4)=(9)/(4) \\ x=(9)/(8) \\ y=x-3 \\ =(9)/(8)-3 \\ y=-(15)/(8) \end{gathered}

The point is (9/8,-15/8)

now we find the distance between the point by using the formula,'


\begin{gathered} \text{length}=\sqrt[]{(y_2-y_1)^2+(^{_{}}x}_2-x_1)^2 \\ \end{gathered}

thus, the distance is,


\begin{gathered} \sqrt[]{((9)/(8)-(1)/(2))^2+(-(15)/(8)-(-(3)/(4)))^2} \\ =\sqrt[]{(53)/(32)} \\ length=1.28 \end{gathered}

Given point P moving along a curve y = x²-1 and point Q moving along a line y = x-example-1
User Ryan Kaskel
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2.8k points