(0, 4) to the hyperbola x2 – 4y2 = 36 are
A. 5x – 6y + 24=0 and 5x + 6y – 24 = 0
B. x – 4y + 16 = 0 and x + 4y – 16 = 0
C. 2x – y + 4 = 0 and 2x + y – 4 = 0
D. None of these
Ans: A
Sol: Any tangent will be in the form y = mx + c, where c2 = 36m2 – 9. If it passes through (0, 4) then 4 = c Þ 16 = 36m2 – 9
m=±5/6
Hence tangent is 6y = 5x + 24 and 6y = –5x + 24
