# At what value of t does the curve x = 2t - 3t^2, y = t^2 - 3t have a vertical tangent?

At what value of t does the curve $$x = 2t - 3t^2, y = t^2 - 3t$$ have a vertical tangent?

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Step 1 The given parametric equations are $$x = 2t − 3t2$$
$$y = t2 − 3t$$ Vertical tangent: It occurs when derivative is undefined for given parametric equations. Step 2 Solve the equations to find derivative with respect to t $$\frac{dx}{dt} = 2 - 6t$$
$${dy}{dt} = 2t - 3$$
$$\frac{dy}{dx} = \frac{dy}{dt} = (\frac{dt}{dx})$$
$$= \frac{2t-3}{2-6t}$$ Now, the derivative is undefined when denominator of the above equation becomes zero i.e. $$2 − 6t = 0$$
$$2 = 6t$$
$$t = \frac{2}{6} = \frac{1}{3}$$ Answer: Hence, for the given parametric equations , the vertical tangent is at $$t = \frac{1}{3}$$ and option (a) is correct answer.

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