The tangent to the hyperbola $xy=c^2$ at the point P intersects the  X-axis at T and the Y-axis at $T \text{'}$. The normal to the hyperbola at P intersects the X-axis at N and the Y-axis at $N\text{'}$; the areas of the $\Delta PNT$ and $\Delta PN\text{'}T\text{'}$are  $\Delta$ and $\Delta \text{'}$ respectively, then $\frac{1}{\Delta }+\frac{1}{\Delta \text{'}}$  

Hyperbola $\Rightarrow xy=c^2$

Let point P(ct,c/t) on the hyperbola

$\therefore$ Equation of tangent at P

$\frac{x}{t}+ty=2c$     …(i)

Normal at P

$y-\frac{c}{t}=x{t}^2-c{t}^3$   …(ii)

Now, for point T and $T\text{'}$ put  $y=0$ and $x=0$ respectively,

We get,  $T=\left(2ct,0\right)$ and $T\text{'}\left(0,2c/t\right)$

Similarly, $N\left(ct-\frac{c}{t^3},0\right)$ and $\left(0,\frac{c}{t}-c{t}^3\right)$

$\Delta =AreaofPNT=\frac{1}{2}\times \frac{c}{t}\times \left[2ct-ct+\frac{c}{t^3}\right)$

$=\frac{c^2}{2t^4}\left(t^4+1\right)$

$\Delta \text{'}=AreaofPN\text{'}T\text{'}=\frac{1}{2}\times ct\left(\frac{2c}{t}-\frac{c}{t}+c{t}^3\right)$

$=\frac{1}{2}{c}^2\left(t^4+1\right)$

$\frac{1}{\Delta }+\frac{1}{\Delta \text{'}}=\frac{2}{c^2}$