The shortest distance between the line $x-y=1$ and the curve $x^2 = 2y$ is:
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To find the shortest distance between a non-intersecting curve and a line, find the point on the curve where the tangent is parallel to the line. The shortest distance is then the perpendicular distance from this point to the line.
The shortest distance between a curve and a line occurs along a common normal. This means the tangent to the curve at the point of shortest distance must be parallel to the given line.
The given line is $x - y = 1$, which can be written as $y = x - 1$. The slope of this line is $m_{line} = 1$.
The given curve is the parabola $x^2 = 2y$, or $y = \frac{1}{2}x^2$.
To find the slope of the tangent to the parabola, we differentiate with respect to x:
$\frac{dy}{dx} = \frac{1}{2}(2x) = x$.
We set the slope of the tangent equal to the slope of the line:
$m_{tangent} = \frac{dy}{dx} = x = 1$.
So, the point on the parabola closest to the line has an x-coordinate of 1.
The corresponding y-coordinate is $y = \frac{1}{2}(1)^2 = \frac{1}{2}$.
The point on the parabola is P(1, 1/2).
Now, we calculate the perpendicular distance from the point P(1, 1/2) to the line $x - y - 1 = 0$.
The formula for the distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.
Here, $(x_0, y_0) = (1, 1/2)$ and the line is $1x - 1y - 1 = 0$.
$d = \frac{|1(1) - 1(1/2) - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|1 - 1/2 - 1|}{\sqrt{1 + 1}} = \frac{|-1/2|}{\sqrt{2}}$.
$d = \frac{1/2}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.
