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?
dy/dx=8t and y=4t^2+c
Then equation of tangent is
but the tangent also passes through (1,0) so its equation can also be
so two equations of same tangent are
so by comparing both we get
so now we also expressed c in terms of t