Conic problem!

Let $Q\left(ct_1, \frac{c}{t_1}\right)$ and $P\left(ct_2, \frac{c}{t_2}\right)$ .

Tangent at $Q$ is given by $\frac{x}{2ct_1}+\frac{y}{\frac{2c}{t_1}}=1$ Tangent at $P$ is given by $\frac{x}{2ct_2}+\frac{y}{\frac{2c}{t_2}}=1$ It now follows, from the first equation and the statement of the question, that $t_2=2t_1$ .

Now, eliminating $t_1$ and $t_2$ from the an above three equations, the required locus comes out to be $xy=\frac{8}{9}c^2$ $\Rightarrow\boxed{9k=8}$