Find the equation of the tangent to the curve y=4x^2+c at the point x=t. If the tangent passes through the point (1,0), find c in terms of t. Help anyone?
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