The shortest distance between the line y - x | vedclass

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(D) The given line is $x - y + 1 = 0$,which has a slope of $1$.For the shortest distance,the tangent to the curve $x = y^2$ must be parallel to the gi

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The given line is $x - y + 1 = 0$,which has a slope of $1$.For the shortest distance,the tangent to the curve $x = y^2$ must be parallel to the given line.Differentiating $x = y^2$ with respect to $y$,we get $\frac{dx}{dy} = 2y$.Since the slope of the tangent is $\frac{dy}{dx} = 1$,we have $\frac{1}{2y} = 1$,which gives $y = \frac{1}{2}$.Substituting $y = \frac{1}{2}$ into $x = y^2$,we get $x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.So,the point $P$ on the curve is $\left(\frac{1}{4}, \frac{1}{2}\right)$.The shortest distance is the perpendicular distance from point $P\left(\frac{1}{4}, \frac{1}{2}\right)$ to the line $x - y + 1 = 0$.Distance $= \frac{|\frac{1}{4} - \frac{1}{2} + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|\frac{3}{4}|}{\sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8}$.

$A$. $P$	$I$. $(2,0)$
$B$. $Q$	$II$. $(\frac{3}{2}, -\frac{1}{4})$
$C$. $S$	$III$. $(\frac{3}{2}, 0)$
$D$. $Z$	$IV$. $(\frac{3}{2}, -\frac{1}{2})$
	$V$. $(0, \frac{3}{2})$

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

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