Question

2. The equation of tangent drawn to the curve y2 - 2x - 4y + 8 =0 from the point (1.2) is given by
(a) y - 2014 J2)=2/3 (x - 2)
(b) y - 2(1+ V3)= 1212 (x - 2)
(c) y - 2 (1£ 13) = + 2/3 (x - 2)
(d) None of these​

Answer 1

Answer:

x²+y²-2x-4y+1 = 0  --------------------(1)

2x+2yy'-2-4y' = 0                 (differentiating)

y' = (1-x)/(y-2)

since tangent is parallel to X axis , y' = 0

1-x = 0 

x = 1

put x = 1 in equation (1)

1+y²-2-4y+1 = 0

y²-4y = 0

y = 0   ,  4

hence points = (1,0)&(1,4)

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Step-by-step explanation: