The shortest distance between line $y-x=1$ and curve $x=y^2$ is
The shortest distance between line $y-x=1$ and curve $x=y^2$ is
-
$\frac{3 \sqrt{2}}{8}$
-
$\frac{8}{3 \sqrt{2}}$
-
$\frac{4}{\sqrt{3}}$
-
$\frac{\sqrt{3}}{4}$
Solution
$
P=\left(y^2, y\right)
$
Perpendicular distance from $P$ to $x-y+1=0$ is $\frac{\left|y^2-y+1\right|}{\sqrt{2}}$
$
\begin{aligned}
& \mathrm{y}^2-\mathrm{y}+1>0 \forall \mathrm{y} \in \mathrm{R} \\
& \therefore \text { Coefficient } \mathrm{y}^2>0 \\
& \therefore \text { Min value }=\frac{1}{\sqrt{2}}\left(\frac{4 \mathrm{ac}-\mathrm{b}^2}{4 \mathrm{a}}\right)=\frac{3}{4 \sqrt{2}}
\end{aligned}
$
Asked in: JEE Main 2011
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