Let the equation of two diameters of a circle x 2 + y 2 - 2 x + 2 f y + 1 = 0 be 2 p x - y = 1 and 2 x + p y = 4 p . Then the slope m ∈ 0 , ∞ of the tangent to the hyperbola 3 x 2 - y 2 = 3 passing through the centre of the circle is equal to _____.

Step-by-step solution

Given, Equation of circle x 2 + y 2 - 2 x + 2 f y + 1 = 0 and given diametric lines will pass through 1 , - f which is the centre of the circle, so 2 p + f - 1 = 0             . . . 1 and 2 - p f - 4 p = 0               . . . 2 Now solving equation 1   &   2 we get, f = 0 or - 3 Now given, Hyperbola 3 x 2 - y 2 = 3   or   x 2 - y 2 3 = 1 We know that tangent to hyperbola is given by y = m x ± m 2 - 3 It passes 1 , 0 &#8658