Question

The normal at the point $P(2,\frac{1}{2})$ on the curve $xy=1$ meets the curve again at $Q.$ If $m$ is the slope of the curve at $Q$, then the value of $|m|$ is

Answer 1

We have $y=\frac{1}{x}$
Slope of tangent, $\frac{dy}{dx}=\frac{-1}{x^2}$
Slope of normal at $P$ is $t^2$


$\therefore t^2=\frac{\frac{1}{t}-\frac{1}{T}}{t-T}=\frac{T-t}{tT}\cdot \frac{1}{t-T}=\frac{-1}{tT}\Rightarrow T=\frac{-1}{t^3}$
Slope of curve at $Q$ is, $m=\frac{-1}{x^2}$
$m=\frac{-1}{T^2}=\left(\frac{-1}{{\left(\frac{-1}{t^3}\right)}^2}\right)=-t^6$

Given that $t=2$
$m=-2^6=-64$
Hence, $|m|=64.$