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Question

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?

Answer 1

y = 4x^2 + c

y&#039; = 8x

The slope at x = t is 8t.

Also, if x = t, then y = 4t^2 + c.

The equation of the line is

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

We know (1,0) is on the line, so

0 - (4t^2 +c) = 8t(1-c)

-4t^2 - c = 8t - 8tc

8tc - c = 4t^2 + 8t

c(8t-1) = 4t^2 +8t

c = (4t^2+8t)/(8t-1)