Let the equation of two diameters of a circle x² + y² –2x + 2fy + 1 = 0 be 2px – y = 1 and 2x + py = 4p. Then the slope $\mathrm{m}\in (0,\mathrm{\infty })$ of the tangent to the hyperbola 3x² –y² = 3 passing through the centre of the circle is equal to _______.

2p + f – 1 = 0                  ........ (1)

2 – pf–4p = 0                 ........ (2)

2 = p(f + 4)

$\mathrm{p}=\frac{2}{\mathrm{f}+4}$

2p = 1 – f

$\frac{4}{\mathrm{f}+4}=1-\mathrm{f}$

f² + 3f = 0

f = 0 or –3

Hyperbola $3{\mathrm{x}}^2-{\mathrm{y}}^2=3,{\mathrm{x}}^2-\frac{{\mathrm{y}}^2}{3}=1$

$\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{m}}^2-3}$

It passes (1, 0)

$\mathrm{o}=\mathrm{m}\pm \sqrt{{\mathrm{m}}^2-3}$

m tends $\infty$

It passes (1, 3)

$\begin{array}{l}3=\mathrm{m}\pm \sqrt{{\mathrm{m}}^2-3}\\ (3-\mathrm{m}{)}^2={\mathrm{m}}^2-3\end{array}$

m = 2