Mathematics - Application of Derivative Question with Solution | TestHub

$P\left(x_0,y_0\right)$ is the nearest point; $y_0=x_0-3$ Line through $P$, perpendicular to $x-y=3$ is normal to given curve at, say, $Q\left(x_1,y_1\right)$ $\therefore -\frac{2}{x_1}=-1;x_1=2;y_1=1.$ Normal is $y-1=-(x-2)$; This cuts $x-y=3$ at $P$. $\therefore P(3,0)$.