Question

If normal to the rectangular hyperbola $xy=4$ at the point $t_1(=4)$ meets the curve again at the point $t_2$, then $\frac{1}{|t_2|}=$

Answer 1

In the rectangular hyperbola $xy=4$, equation of normal at $t_1$ is
$(y-\frac{2}{t_1})=t_1^2(x-2t_1)\Rightarrow y-\frac{1}{2}=16(x-8)\cdots \left(1\right)$
Here $\left(1\right)$ passes through $(2t_2,\frac{2}{t_2})$
$\Rightarrow (\frac{2}{t_2}-\frac{1}{2})=16(2t_2-8)$
$\Rightarrow \frac{4-t_2}{2t_2}=32(t_2-4)$
$\Rightarrow t_2=-\frac{1}{64}[\because t_1\ne t_2]$