Question

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 $m \in(0, \infty)$ of the tangent to the hyperbola $3 x^2-y^2=3$ passing through the centre of the circle is equal to _____

Answer 1

$ 2 p + f -1=0 .....$(1)
$2-p f -4 p =0 .....$(2)
$ 2=p( f +4) $
$ p =\frac{2}{ f +4} $
$ 2 p =1- f $
$ \frac{4}{ f +4}=1- f $
$ f ^2+3 f =0$
$ f =0 \text { or }-3$
Hyperbola $3 x^2-y^2=3, x^2-\frac{y^2}{3}=1$
$y = mx \pm \sqrt{ m ^2-3}$
It passes $(1,0)$
$o = m \pm \sqrt{ m ^2-3}$
$m$ tends $\infty$
It passes $(1,3)$
$ 3=m \pm \sqrt{m^2-3}$
$ (3-m)^2=m^2-3$
$m=2$