Question

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.

Answer 1

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,

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

thus,

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

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,

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

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

$\begin{gathered} -x-\frac{1}{4}=x-3 \\ 2-x=3-\frac{3}{4}=2\frac{1}{4}=\frac{9}{4} \\ x=\frac{9}{8} \\ y=x-3 \\ =\frac{9}{8}-3 \\ y=-\frac{15}{8} \end{gathered}$

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,

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