Question

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 m ∈ (0, ∞) of the tangent to the hyperbola 3x2 – y2 = 3 passing through the center of the circle is equal to _______.

Answer 1

2p + f – 1 = 0 ........ (1)
2 – pf–4p = 0 ........ (2)
2 = p(f + 4)
p=$\frac{2}{f+4}$
2p = 1 – f
$\frac{f}{f+4}$=1−f
f2 + 3f = 0
f = 0 or –3
Hyperbola $3x^2−y^2$=3,$\frac{x^2−y^2}{3}$=1
y=mx±$\sqrt{m^2-3}$
It passes (1, 0) o=m±$\sqrt{m^2-3}$ ,m tends ∞
It passes (1, 3)
3=m±$\sqrt{m^2-3}$ (3−m)2=m2−3
m = 2