Solution :
Ensure first that point(1,2) doesn't lie on a hyperbola by substituting (1,2) on hyperbola.
1^2/4-2^2/16=1/4-4/16=1/4-1/4=0 differs from 1.
Thus this point (1,2) doesn't lie on hyperbola
Power line of point(1,2) to the hyperbola x^2/4-y^2/16=1
x1*x/4-y1*y/16=1
where (x1,y1)=(1,2)
1*x/4-2*y/16=1
x/4-y/8=1
2x-y=8
y=2x-8=2(x-4)
substitute to hyperbola
x^2/4-2^2*(x-4)^2/16=1
x^2/4-(x-4)^2/4=1
x^2-(x-4)^2=4
x^2-x^2+8x-16=4
8x=20
x=2.5
substitute to hyperbola
(25/4)/4-y^2/16=1
25/16-y^2/16=1
25-y^2=16
y^2-9=0
(y-3)*(y+3)=0
y-3=0
y1=3
y+3=0
y2=-3
the cross connect points between power line and hyperbola are (2.5,3) and (2.5,-3)
The tangent lines found by connecting point(1,2) to (2.5,3) and (2.5,-3)
Connecting point(1,2) to (2.5,3)
(y-y1)/(y1-y2)=(x-x1)/(x1-x2)
where(x1,y1)=(1,2)
(x2,y2)=(2.5,3)
(y-2)/(2-3)=(x-1)/(1-2.5)
(y-2)/(-1)=(x-1)/(-1.5)
3*(y-2)=2*(x-1)
3y-6=2x-2
2x-3y+4=0 [1]
Connecting point(1,2) and(2.5,-3)
(y-2)/(2+3)=(x-1)/(1-2.5)
(y-2)/5=(x-1)/(-1.5)
-3*(y-2)=10*(x-1)
-3y+6=10x-10
10x+3y-16=0 [2]
[1] and [2] are the tangent lines to be solved