If x = t^2 and y = 2t,what is the equation of | vedclass

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(D) Given the parametric equations $x = t^2$ and $y = 2t$. First,find the derivative $\frac{dy}{dx}$: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$. Th

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$x + y - 3 = 0$

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(D) Given the parametric equations $x = t^2$ and $y = 2t$. First,find the derivative $\frac{dy}{dx}$: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$. Therefore,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{2t} = \frac{1}{t}$. At $t = 1$,the slope of the tangent is $m = \frac{1}{1} = 1$. The slope of the normal is $m' = -\frac{1}{m} = -1$. At $t = 1$,the coordinates of the point are $x = (1)^2 = 1$ and $y = 2(1) = 2$. The equation of the normal line passing through $(1, 2)$ with slope $-1$ is: $y - 2 = -1(x - 1)$ $y - 2 = -x + 1$ $x + y - 3 = 0$.

If $x = t^2$ and $y = 2t$,what is the equation of the normal at $t = 1$?

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