Question

Let normal at point $P(1,2)$ on the curve $xy=2$ meets the curve again at point $Q,$ then the value of $\frac{4P{Q}^2}{5}=$

Answer 1

Let the point $Q\equiv (t,\frac{2}{t})$
For slope of tangent , $y=\frac{2}{x}$
$\Rightarrow \frac{dy}{dx}=-\frac{2}{x^2}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(1,2)}=-2$
$\Rightarrow \text{slope of normal}=-\frac{1}{\text{slope of tangent}}$
$\Rightarrow$ slope of normal =$\frac{1}{2}$=slope of PQ
$\Rightarrow \frac{1}{2}=\frac{\frac{2}{t}-2}{t-1}$
$\Rightarrow t^2+3t-4=0$
$\Rightarrow (t+4)(t-1)=0$
$\Rightarrow t=-4,1$
$\therefore Q\equiv (-4,-\frac{1}{2}),p\equiv (1,2)$
$P{Q}^2=\frac{125}{4}$
$\therefore \frac{4P{Q}^2}{5}=25$