Thanks for the reply. I'm very tired and I've got to get up early, so I've not really got time to think about the second part, but here's a starting point for you to investigate:
The gradient of the tangent y = 2x + K = 2
So that means that you're seeking a point on the curve where the gradient is also 2. Hence dy/dx must be equal to 2.
Actually, looking at it again, it might be better to find dx/dy (which must be equal to the reciprocal of 2, i.e. ½)
x =1/8 y^2
gives dx/dy = ¼y
<=> ¼y = ½
<=> y = 2
Since y = 2 and, on the curve, 8x = y^2
<=> x = ½
Substituting into y = 2x + K we get
2 = 1 + K
<=> K = 1
Oh, I think I might have done it anyway - but that working is UNCHECKED. It might turn out to be RUBBISH! I'll leave it to you to check it!
Chris