Question

Let the equation of two diameters of a circle x 2 + y 2 - 2 x + 2 fy + 1 = 0 be 2 px - y = 1 and 2 x + py = 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 _______.

Answer 1

x 2 + y 2 – 2 x + 2 fy + 1 = 0 [entre = (1, – f]
Diameter 2 px – y = 1 …(i)
2 x + py = 4 p …(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