The shortest distance between the line x-y=1 | vedclass

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(B) The shortest distance between a curve and a line occurs at a point $P(x_0, y_0)$ on the curve where the tangent is parallel to the given line.The

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$\frac{1}{2\sqrt{2}}$

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(B) The shortest distance between a curve and a line occurs at a point $P(x_0, y_0)$ on the curve where the tangent is parallel to the given line.The equation of the line is $x-y=1$,which can be written as $y=x-1$. The slope of this line is $m=1$.The equation of the curve is $x^2=2y$,which implies $y=\frac{x^2}{2}$.The slope of the tangent to the curve at any point $(x_0, y_0)$ is given by $\frac{dy}{dx} = \frac{d}{dx}(\frac{x^2}{2}) = x$.Setting the slope of the tangent equal to the slope of the line,we get $x_0 = 1$.Substituting $x_0=1$ into the curve equation $y_0 = \frac{x_0^2}{2}$,we get $y_0 = \frac{1^2}{2} = \frac{1}{2}$.Thus,the point on the curve is $P(1, \frac{1}{2})$.The shortest distance from a point $(x_1, y_1)$ to the line $Ax+By+C=0$ is given by $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$.Here,the line is $x-y-1=0$,so $A=1, B=-1, C=-1$. The point is $(1, \frac{1}{2})$.Shortest distance $= \left|\frac{1(1) + (-1)(\frac{1}{2}) - 1}{\sqrt{1^2+(-1)^2}}\right| = \left|\frac{1 - \frac{1}{2} - 1}{\sqrt{2}}\right| = \left|\frac{-\frac{1}{2}}{\sqrt{2}}\right| = \frac{1}{2\sqrt{2}}$.

The shortest distance between the line $x-y=1$ and the curve $x^{2}=2y$ is .... .

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