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

Correct Answer: 2

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