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

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Question

(A) The given line is $x-y+1=0$. The curve is $x=y^2$. Let a point on the curve be $P(y^2, y)$. The slope of the tangent to the curve at $P$ must be e

Vedclass · Accepted Answer

$\frac{3 \sqrt{2}}{8}$

Vedclass · Answer

(A) The given line is $x-y+1=0$. The curve is $x=y^2$. Let a point on the curve be $P(y^2, y)$. The slope of the tangent to the curve at $P$ must be equal to the slope of the given line for the shortest distance. The slope of the line $x-y+1=0$ is $m=1$. Differentiating $x=y^2$ with respect to $y$,we get $\frac{dx}{dy} = 2y$. Thus,$\frac{dy}{dx} = \frac{1}{2y}$. Setting $\frac{1}{2y} = 1$,we find $y = \frac{1}{2}$. Then $x = y^2 = (\frac{1}{2})^2 = \frac{1}{4}$. The point on the curve is $P(\frac{1}{4}, \frac{1}{2})$. The shortest distance $d$ from point $(x_0, y_0)$ to the line $Ax+By+C=0$ is given by $d = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$. Here,$A=1, B=-1, C=1, x_0=\frac{1}{4}, y_0=\frac{1}{2}$. $d = \frac{|1(\frac{1}{4}) - 1(\frac{1}{2}) + 1|}{\sqrt{1^2+(-1)^2}} = \frac{|\frac{1}{4} - \frac{2}{4} + \frac{4}{4}|}{\sqrt{2}} = \frac{3/4}{\sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8}$.

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

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