y=4x^2+c

dy/dx=8x

at x=t

dy/dx=8t and y=4t^2+c

Then equation of tangent is

y-(4t^2+c)=8t(x-t)

y=4t^2+c+8tx-8t^2 or

y=8tx-4t^2+c

but the tangent also passes through (1,0) so its equation can also be

y-0=8t(x-1) or

y=8tx-8t

so two equations of same tangent are

1)y=8tx-4t^2+c

2)y=8tx-8t

so by comparing both we get

-8t=-4t^2+c

or

c=4t^2-8t

so now we also expressed c in terms of t