If x = t^2 and y = 2t,then the equation of th | vedclass

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(A) Given $x = t^2$ and $y = 2t$. At $t = 1$,$x = 1^2 = 1$ and $y = 2(1) = 2$. So,the point is $(1, 2)$. Now,$\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} =

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$x + y - 3 = 0$

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(A) Given $x = t^2$ and $y = 2t$. At $t = 1$,$x = 1^2 = 1$ and $y = 2(1) = 2$. So,the point is $(1, 2)$. Now,$\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$. 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 $m' = -\frac{1}{m} = -\frac{1}{1} = -1$. The equation of the normal at $(1, 2)$ is $y - y_1 = m'(x - x_1)$. $y - 2 = -1(x - 1)$. $y - 2 = -x + 1$. $x + y - 3 = 0$.

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

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