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(A) Given parametric equations are $x=t^2$ and $y=2t$. First,find the derivatives: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$. The slope of the tang

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$2x+y-12=0$

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(A) Given parametric equations are $x=t^2$ and $y=2t$. First,find the derivatives: $\frac{dx}{dt} = 2t$ 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 $t=2$,the slope of the tangent is $m_T = \frac{1}{2}$. The slope of the normal is $m_N = -\frac{1}{m_T} = -2$. At $t=2$,the coordinates of the point are $x = (2)^2 = 4$ and $y = 2(2) = 4$. The equation of the normal at $(4, 4)$ with slope $m_N = -2$ is given by $(y - y_1) = m_N(x - x_1)$. Substituting the values: $(y - 4) = -2(x - 4)$. $y - 4 = -2x + 8$. $2x + y - 12 = 0$.

If $x=t^2$ and $y=2t$ are parametric equations of a curve,then the equation of the normal to the curve at $t=2$ is

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