S1: Centre of the hyperbola x2 – 4y2 – 4x + 8y + 4 = 0 is (2,1)
S2: If eccentricity of hyperbola x(y – 1) = 2 is √2 then eccentricity of its conjugate hyperbola is 2.
S3: From point (2, 2) tangents are drawn to the hyperbola \(\frac{x^2}{16} - \frac{y^2}9=1\), then point of contact lie in I & IV quadrants.
S4: Product of the length of perpendiculars drawn from any foci of the hyperbola x2 – 4y2 – 4x + 8y + 4 = 0 to its asymptotes is 4 .
(a) TFTT
(b) TFFT
(c) TTFT
(d) TTTT