What is the equation of the tangent to the cu | vedclass

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(C) Given the parametric equations of the curve are $x = at^2$ and $y = 2at$.First,we find the derivative $\frac{dy}{dx}$:$\frac{dx}{dt} = 2at$ and $\

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$ty = x + at^2$

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(C) Given the parametric equations of the curve are $x = at^2$ and $y = 2at$.First,we find the derivative $\frac{dy}{dx}$:$\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$.Therefore,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$.The equation of the tangent at point $(at^2, 2at)$ is given by $(y - y_1) = m(x - x_1)$.Substituting the values: $(y - 2at) = \frac{1}{t}(x - at^2)$.Multiplying both sides by $t$,we get $ty - 2at^2 = x - at^2$.Rearranging the terms,we get $ty = x + at^2$.

What is the equation of the tangent to the curve $x = at^2, y = 2at$ at any point $t$?

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