A maths

1. P(4,12), Q(1,6)
PQ : (y-6)/(x-1) = (12-6)/(4-1) = 2
PQ : y = 2x + 4

2.
dy/dt = -12/(2+t)^2
dx/dt = -8/(2+t)^3
dy/dx = 3(2+t)/2
Slope of PQ = 2
Normal is perpendicular to PQ
dy/dx = 2
3(2+t)/2 = 2
t = -2/3
Point of contact = (9/4,9)
y-9 = 2(x - 9/4)
y = 2x + 9/2
2y = 4x + 9

1)
When t = -1, x = 4 & y = 12
P = (4,12)
When t = 0, x = 1 & y = 6
Q = (1,6)
So, equation of PQ
y-6 = [(6-12)/(1-4)](x-1)
y-6 = [-6/-3](x-1)
y-6 = 2(x-1)
y = 2x+4
*******
2)
y = 12/ 2+t
y^2 = 144/(2+t)^2
y^2 = 36x is the normal form of the curve.
Differentiate wrt x
2y(y') = 36
y' = 18/y
To let y' = 18/y = slope 2
we get y = 9, then x = 81/36 = 9/4
So, the equation of the required normal to the curve is
[y-9 / x-(9/4)] = -1/2
x+2y-81/4 = 0
4x+8y-81 = 0