Re: HSC 2015 2U Marathon
$Since the point $P(x_1,y_1)$ lies on $y=\sqrt{x-4}$, it can be expressed as $P(x_1,\sqrt{x_1-4})$. Using $m=\frac{y_2-y_1}{x_2-x_1}$, the gradient of $OP$ is $m=\frac{\sqrt{x_1-4}}{x_1}$. Since $OP$ is the tangent to $y=\sqrt{x-4}$ at $x=x_1$, you can differentiate...