The parametric equations of a curve are x=2+3t^2, y=2t^3. Find the equation of the tangent to the curve at the point with parameter t. I managed to solve it and answer that I got is y=tx-2t-t^3.
Hence, show that all tangents to the curve except x-axis cut the x-axis at the points where x>2,
what does it mean?how can I prove it? thanks for your guides!
Hence, show that all tangents to the curve except x-axis cut the x-axis at the points where x>2,
what does it mean?how can I prove it? thanks for your guides!
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dx/dt = 6t and dy/dt = 6t^2
=> dy/dx = (6t)/(6t) = t
=> eqn. of tangent is y - 2t^3 = t (x - 2 - 3t^2)
=> y = tx - 2t - t^3 ... [You are right]
plugging y = 0 for x-axis
tx - 2t - t^3 = 0
=> t (x - 2 - t^2) = 0
Since x-axis is to be excluded, (t ≠ 0, because t = 0 corresponds to x-axis)
=> x = 2 + t^2
=> x > 2 ... [because t^2 > 0. t cannot be zero as it corresponds to x-axis].
=> dy/dx = (6t)/(6t) = t
=> eqn. of tangent is y - 2t^3 = t (x - 2 - 3t^2)
=> y = tx - 2t - t^3 ... [You are right]
plugging y = 0 for x-axis
tx - 2t - t^3 = 0
=> t (x - 2 - t^2) = 0
Since x-axis is to be excluded, (t ≠ 0, because t = 0 corresponds to x-axis)
=> x = 2 + t^2
=> x > 2 ... [because t^2 > 0. t cannot be zero as it corresponds to x-axis].
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I assume you meant x ≥ 2.....set the tangent y = 0 and solve for x....x = 2 + t².....ahhh..
except for a point on the x axis { where t = 0 }
except for a point on the x axis { where t = 0 }