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 centre of the circle is equal to _______.

Answer 1

Correct Answer: 2

x² + y² – 2x + 2fy + 1 = 0 [entre = (1, –f]
Diameter 2px – y = 1 …(i)
2x + py = 4p …(ii)
x=\(\frac{5P}{2P^2+2}\)
y=\(\frac{4P^2-1}{1+P^2}\)
∵x=1
f = 0
[for P=\(\frac{1}{2}\)]
\(\frac{5P}{2P^2+2}\)=1
f = 3 [for P = 2]
∴P=\(\frac{1}{2}\),2
Centre can be(12,0) or (1,3)
(12,0)will not satisfy
∴ Tangent should pass through (2, 3) for 3x² – y² = 3
\(\frac{x^2}{1}\)−\(\frac{y^2}{3}\)=1
\(y=mx\pm\sqrt{m^2-3}\)
Substitute (2, 3)
3=\(mx\pm\sqrt{m^2-3}\)
∴ m=2